Quadrilaterals And Parallelograms

Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360°, we have:
∴ 56° + 115° + 84° + x° = 360°
∴ 255° + x° = 360° ​
∴ x° = 105°
Hence, the measure of the fourth angle is 105°.

Our given ratio of angles is 2:4:5:7. Let common multiplying factor be x°.

Hence, ∠A = 2x°, ∠B = 4x°, ∠C = 5x° and ∠D = 7x°
Since the sum of the angles of a quadrilateral is 360°, we have:
∴ 2x + 4x + 5x + 7x = 360°

∴ 18 x = 360° ​
∴ x = 20°
∴ ∠A = 40°; ∠B = 80°; ∠C = 100°; ∠D = 140°

Hence, the measure of the angles are 40°, 80°, 100° and 140°

Here given that ABCD is trapezium where AB || DC.

We observe that ∠ A and ∠ D are the interior angles on the same side of transversal line AD, whereas ∠ B and ∠ C are the interior angles on the same side of transversal line BC.

As ∠A and ∠D are interior angles, we have,

∠A + ∠D = 180°
∴ ∠D = 180° − ∠A
∴ ∠ D = 180° − 55° = 125°
Similarly for ∠ B and ∠C,

∠ B + ∠C = 180°
∴ ∠C = 180°− ∠B
∴ ∠ C = 180° − 70° = 110°
Hence, measure of ∠ D and ∠ C are 125°and 110° respectively.

(i) Here it is given that in ABCD is a square and ΔEDC is an equilateral triangle.

Hence, we say that AB = BC = CD = DA and ED = EC = DC

Now in ∆ADE and ∆BCE, we have,
AD = BC … given

DE = EC​ … given
∠ADE = ∠BCE … as both angles are sum of 60° and 90°

∴ ∆ADE≅ ∆BCE

Now by cpct,

AE = BE …(1)

(ii) Here ∠ADE = 90°+ 60° = 150°​

DA = DC … given
DC = DE … given

∴ DA = DE

This means that sides of square and triangles are equal.

∴ ∆ADE and ∆BCE are isosceles triangles.

Hence, ∠DAE = ∠DEA = (180° − 150°) = 30°/2 =15°

Given: In ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.

To prove: bisects BD ie. DO = BO

Proof:

Now, in ∆OND and ∆OMB, we have,
∠OND = ∠OMB …90° each
∠DON = ∠ BOM …Vertically opposite angles
Also, DN = BM …Given
Hence, by AAS congruence rule,

∆OND ≅ ∆OMB
∴ OD = OB …CPCT
​Hence, AC bisects BD.

Given: In ABCD, and

To prove: (i) bisects and

(ii)

(iii)

Proof:

(i) In ∆ABC and ∆ADC, we have,
AB = AD …given

BC = DC …given
AC = AC … common side
Hence, by SSS congruence rule,

∆ABC ≅ ∆ADC

∴ ∠BAC = ∠DAC and ∠BCA = ∠D​CA …By cpct
Thus, AC bisects ∠A and ∠ C.
(ii) Now, in ∆ABE and ∆ADE, we have,

AB = AD …given​
∠BAE = ∠DAE​ …from i
AE = AE …common side
Hence, by SAS congruence rule,

∆ABE ≅ ∆ADE
∴ BE = DE …by cpct
(iii) ∆ABC ≅ ∆ADC from ii

∴ ∠ABC = ∠AD​C …by cpct

Given: ABCD is where and

To prove: (i) (ii)

(iii)

Proof:

(i) Here,

BC = CD …Sides of square

CQ = DR …Given

BC = BQ + CQ

∴ CQ = BC − BQ

∴ DR = BC – BQ ...(1)

Also,

CD = RC+ DR

∴ DR = CD − RC = BC − RC ...(2)

From (1) and (2), we have,

BC − BQ = BC − RC

∴ BQ = RC

(ii) Now in ∆RCQ and ∆QBP, we have,

PB = QC …Given

BQ = RC …from (i)

∠RCQ = ∠QBP …90° each

Hence by SAS congruence rule,

∆RCQ ≅ ∆QBP

∴ QR = PQ …by cpct

(iii) ∆RCQ ≅ ∆QBP and QR = PQ … from (ii)

∴ In ∆RPQ,

∠QPR = ∠QRP = (180° − 90°) = = 45°

∴ ∠QPR = 45°

Given: In ABCD, O is any point within the quadrilateral.
To prove:
Proof:

We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC,

OA + OC > AC …(1)

Also, in ∆ BOD,

OB + OD > BD …(2)

Adding 1 and 2, we get,

(OA + OC) + (OB + OD) > (AC + BD)
∴ OA + OB + OC + OD > AC + BD

Hence proved.

Given: In ABCD, AC is one of diagonals.

To prove:

(i)

(ii)

(iii)

Proof:

(i) We know that the sum of any two sides of a triangle is greater than the third side.
In ∆ABC,

AB + BC > AC ...(1)

In ∆ACD,

CD + DA > AC ...(2)

​Adding (1) and (2), we get,

AB + BC + CD + DA > 2AC

(ii) In ∆ABC, we have,
AB + BC > AC ...(1)
We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
In ∆ACD, we have:​
AC > DA – CD ...(2)
From (1) and (2), we have,
AB + BC > DA − CD
∴ AB + BC + CD > DA

(ii) In ∆ABC,

AB + BC > AC …(1)

In ∆ACD,

CD + DA > AC …(2)

In ∆ BCD,

BC + CD > BD …(3)

In ∆ ABD,

DA + AB > BD …(4)
​Adding 1, 2, 3 and 4, we get,
2(AB + BC + CD + DA) > 2(AC + BD)
∴ AB + BC + CD + DA > AC + BD

Given: Consider a PQRS where QS is diagonal.

To prove: ∠P + ∠Q + ∠R + ∠S = 360°

Proof:

For ∆PQS, we have,

∠P + ∠PQS + ∠PSQ = 180° ... (1) …Using Angle sum property of Triangle

Similarly, in ∆QRS, we have,

∴ ∠SQR + ∠R + ∠QSR = 180° ... (2) …Using Angle sum property of Triangle

On adding (1) and (2), we get

∠P + ∠PQS + ∠PSQ + ∠SQR + ∠R + ∠QSR = 180° + 180°

∴ ∠P + ∠PQS + ∠SQR + ∠R + ∠QSR + ∠PSQ = 360°

∴ ∠P + ∠Q + ∠R + ∠S = 360°

∴ The sum of all the angles of a quadrilateral is 360°.

In ABCD, ∠A = 72°

We know that opposite angles of a parallelogram are equal.
Hence, ∠A ​= ∠C and ∠B ​= ∠D ​ ​
∴ ∠C = 72°
∠A and ∠B are adjacent angles.
∴ ∠A ​+ ∠B​ = 180°
∠B = 180°° − ∠A
​ ∠B​ = 180° − 72° = 108°
∴​ ∠B​ =​ ∠D = 108°
Hence, ∠B​ =​ ∠D = 108°​ and ∠C​ = 72°

​

It is given that ABCD is parallelogram and ∠DAB = 80°​ and ∠DBC = 60°
We need to find measure of ∠CDB and ∠ADB
In ABCD, AD ||​BC, BD as transversal,
∠DBC = ∠ ADB = 60° …Alternate interior angles ...(i)
As ∠DAB and ∠ADC are adjacent angles,

∠DAB ​+ ∠ADC​ = 180°
∴ ∠ADC = 180°° − ∠DAB
​∠ADC​ = 180° − 80° = 100°
Also,

∠ADC​ = ∠ADB + ∠C​​DB
∴​ ∠ADC​ =​ 100°
∠ADB + ∠C​​DB = 100° ...(ii)
From (i) and (ii), we get:
60° + ∠C​​DB = 100°
⇒ ∠C​​DB = 100° − 60° = 40°
Hence, ∠CDB​ =​ 40° and ∠ADB​ = 60°​

Given: ABCD is a parallelogram. The bisectors of and meet at .
To prove: (i) (ii) and (iii)

Proof:

∴ ​∠A = ​∠C and ∠B =​ ∠D … Opposite angles
And ​∠A + ​∠B = 180° … Adjacent angles
∴ ​∠B = 180° − ∠A
180° − 60° = 120° … as ∠A = 60°
∴ ​∠A = ​∠C = 60° and ∠B =​ ∠D​ = 120°
(i) In ∆ APB,

∠​PAB = = 30°and ∠PBA = = 60°
∴​ ∠​APB​ = 180° − (30° + 60°) = 90°
(ii) In ∆ ADP, ∠​PAD = 30° and ∠ADP = 120°
∴ ∠APB = 180° − (30° + 120°) = 30°

Hence,

∠​PAD = ​∠APB = ​30°
Hence, ∆ADP is an isosceles triangle and AD = DP.
In ∆ PBC,

∠​ PBC = 60°

∠​ BPC = 180° − (90° +30°) = 60°and∠​ BCP = 60° …Opposite angle of ∠A
∴ ∠ PBC = ∠​ BPC = ∠​ BCP
Hence, ∆PBC is an equilateral triangle and, therefore, PB = PC = BC.​
(iii) DC = DP + PC
From (ii), we have

DC = AD + BC …AD = BC
DC = AD + AD

DC = 2 AD

In ABCD, and

(i) In ∆AOB,

∠BAO = 35°

∠AOB = ∠COD = 105° …Vertically opposite angels
∴ ​∠ABO = 180° − (35° + 105°) = 40° … Using Angle sum property of Triangle
(ii) ∠ODC and ∠ABO are alternate angles for transversal BD
∴ ∠ODC = ∠ABO = 40°
(iii) ∠ACB = ∠​CAD = 40°° …Alternate angles for transversal AC
(iv) ∠CBD = ∠ABC − ∠ABD ...(1)

∠ABC = 180° − ∠BAD …Adjacent angles are supplementary

​∠ABC = 180° − 75° = 105°
∠CBD = 105° − ∠ABD … as ∠ABD = ∠ABO
∠CBD = 105° − 40° = 65°

It is given that in ABCD, and
We know that opposite angles of parallelogram are equal.

∴∠A = ∠C and ∠B = ∠D
Also,

∠A + ∠B = 180° …Adjacent angles of parallelogram are supplementary
∴​ (2x + 25)°​ + (3x − 5)°​ = 180°
​5x° +20° = 180°
​ 5x° = 160°
​ x° = 32°
∴​∠A = 2 ⨯ 32 + 25 = 89°

∴ ∠B = 3 ⨯ 32 − 5 = 91°
Hence, x = 32°, ∠A = ∠C = 89° and ∠B = ∠D = 91°

Let ABCD be the parallelogram.

We know that opposite angles of parallelogram are equal.

∴∠A = ∠C and ∠B = ∠D
By given conditions,

Let ∠A = x° and ∠B =

Also, adjacent angles of parallelogram are supplementary,

∴ x° + = 180°

= 180°

∴ x = 100°

Hence, ∠A = 100° and ∠B = = 80°

Hence, ∠A = ∠C = 100°; ∠B = ∠D = 80°

Let ABCD be the parallelogram.

We know that opposite angles of parallelogram are equal.

∴∠A = ∠C and ∠B = ∠D
Let ∠A be the smallest angle whose measure is x°.
∴​ ∠​B = (2x − 30)°
We know that adjacent angles of parallelogram are supplementary,

∠​A + ∠B = 180°
x + 2x − 30° = 180°
3x = 210°
x = 70°
∴ ​∠​B = 2 ⨯ 70° − 30° = 110°
Hence, ∠A = ∠C = 70° and ∠B = ∠D = 110°

Here ABCD is parallelogram.

We know that the opposite sides of a parallelogram are parallel and equal.

Hence, AB = DC = 9.5 cm

Also let BC = AD = x cm

Now,

Perimeter of ABCD = 30 cm …(given)

∴ AB + BC + CD + DA = 30 cm

∴ 9.5 + x + 9.5 + x = 30

∴ 19 + 2x = 30
∴ 2x = 11
∴ x = 5.5 cm

Hence, length of each side is AB = DC = 9.5 cm and BC = DA = 5.5 cm

(i) ABCD is a rhombus.

We know that rhombus is type of parallelogram whose all sides are equal.

In ∆ABC, ∠​BAC = ∠BCA = (180° − 110°) = 35°

Hence x = 35°

But AB || DC …opposite sides of rhombus are parallel

∠​BAC = ∠DCA …for transversal AC

∴ ∠​BAC = ∠DCA = 35°

Hence, x = y = 35°

(ii) ABCD is a rhombus.

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

∴ in ∆​AOB, ​

∠OAB = 40°, ∠AOB = 90°

∴ ∠ABO = 180° − (40° + 90°) = 50°

Hence x = 50°

Now in ∆DAB,

AB = AD … as rhombus has all sides equal.

ie. ∆​AOB is isosceles triangle.

Also base angles of isosceles triangle are equal.

Hence, x = y = 50°

(iii) ABCD is a rhombus.

We know that rhombus is type of parallelogram whose all sides are equal.

So in ∆DCB,

DC = BC

∴ ∠​CDB = ∠CBD = y° base angles of isosceles triangle are equal.

Now, x = ∠​CAB …alternate angles with transversal AC

∴ x = ∠​BAD

∴x =× 62°

∴x = 31°

In ∆DOC,

We know sum of angles of triangle is 180°

∠CDO + ∠DOC + ∠OCD = 180°

∴ ∠CDO + 90° + 31° = 180°

∴ ∠CDO = 59°

∴ y = 59°

Hence, x = 31° and y = 59°

Let ABCD be rhombus.

Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle in which OA = = 12 cm and OB = = 9 cm.
Now, AB²= OA² + OB² …Pythagoras theorem
∴​ AB²=​ (12)² + (9)²
∴​ AB²=​ 144 + 81 = 225
∴​ AB =​ 15 cm
Hence, the side of the rhombus is 15 cm

Let ABCD be rhombus.

We know that rhombus is type of parallelogram whose all sides are equal.

∴ AB = BC = CD = DA = 10 cm

Let the diagonals AC and BD intersect each other at O, where AC = 16 cm and let BD = x
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴​ ∆AOB is a right angle triangle, in which OB = BD ÷ 2 = x ÷ 2 and OA = AC ÷2 = 16 ÷ 2 = 8 cm.
Now, AB= OA² + OB²…by pythagoras theorem
∴ 10² = ( )² + 8²

ie. 100 − 64 =

36 ×4 = x²
∴ x² =144
∴ x = 12 cm

Hence, the length of the other diagonal is 12 cm

We know that area of rhombus is,

Area of rhombus = × ( Diagonal1) × ( Diagonal2)

Hence,

Area of ABCD = × AC × BD

= × 16 × 12

= 96 cm²

Hence, the area of rhombus is 96 cm²

(i) Here, ABCD is rectangle.

We know that the diagonals of a rectangle are congruent and bisect each other.

∴ In ∆ AOB, we have OA = OB

This means that ∆ AOB is isosceles triangle.

We know that base angles of isosceles triangle are equal.

∴ ∠​OAB = ∠​OBA = 35°

∴ ∴​ x = 90° − 35° = 55°

Also, ∠AOB = 180° − (35° + 35°) = 110°

∴ y = ∠AOB​ = 110°​ …Vertically opposite angles

Hence, x = 55° and y = 110°​​

(ii) Here, ABCD is rectangle.

We know that the diagonals of a rectangle are congruent and bisect each other.

∴ In ∆ AOB, we have OA = OB

This means that ∆ AOB is isosceles triangle.

We know that base angles of isosceles triangle are equal.

∴ ∠​OAB = ∠​OBA =× (180° − 110°) = 35°

∴​ y = ∠BAC = 35° … alternate angles with transversal AC
Also, x = 90° – y … ​∵∠C = 90° = x + y
∴​ x = 90° − 35° = 55°
Hence, x = 55° and y = 35°​​

Here, ABCD is square.

Here AC and BD are diagonals.

We know that the angles of a square are bisected by the diagonals.

∴ ∠​OBX = 45° ∵∠​ABC = 90° and BD bisects ∠​ABC​
And ∠​BOX = ∠COD = 80° … Vertically opposite angles
∴​ In ∆BOX, we have:
∠AXO =∠OBX + ​∠BOX … Exterior angle theorem
⇒​ ∠AXO = 45° + 80° = 125°
∴ ​x =125°

Here, ABCD is parallelogram.

Hence, AD || BC and AD = BC

(i) In ∆ALD and ∆CMB, we have,
AD = BC

∠​ALD = ∠​CMB (90^o each)

∠​ADL = ∠​CBM (Alternate interior angle)
∴ ∆ALD ≅ ∆CMB

(ii) As ∆ALD ≅ ∆CMB …from 1
∴​ AL = CM …by cpct

ABCD is parallelogram.

We know that the sum of the adjacent angles in parallelogram is 180°

∴ ​∠ A + ∠B = 180°

∴ + = = 90°

In ∆​ APB, we have:
∠​PAB = ∠A​ /2
∠​PBA = ∠B​ /2​
∴ ∠​APB = 180 − (∠​PAB​ + ∠​PBA) …Angle sum property of triangle
∴ ​∠APB = 180 – ( + )
∴ ∠APB = 180 − 90 = 90°
Hence, proved.

ABCD is parallelogram

We know that opposite sides and angles of parallelogram are equal.

∴ ∠B = ∠D and AD = BC and AB = DC

Also,AD || BC and AB|| DC

It is given that and

Hence, ​AP = CQ … ∵ AD = BC

In ∆​DPC and ∆​BQA, we have,

AB = CD

∠B = ∠D

DP = QB … as and

Hence, by SAS test for congruency,

∆​DPC ≅ ∆​BQA

∴PC = QA … by cpct

Hence, from above, in AQCP, we have,

AP = CQ and PC = QA

∴ AQCP is a parallelogram.

ABCD is parallelogram.

∴ in ∆​ODF and ∆​OBE, we have:

OD = OB … Diagonals bisects each other
∠DOF = ∠BOE … Vertically opposite angles
∠FDO = ∠OBE … Alternate interior angles

Hence, by SAA test for congruency,
∆​ODF ≅ ∆​OBE
∴​ OF = OE …by cpct
Hence, proved.

ABCD is parallelogram.

In ∆ODC and ∆​OEB, we have,
DC = BE …as DC = AB
∠COD = ∠BOE … Vertically opposite angles are equal
∠OCD = ∠OBE … Alternate angles with transversal BC
Hence, by SAA test for congruency, we get,
∆​ODC ≅ ∆​OEB

∴ OC = OB …by cpct
We know that BC = OC + OB.
∴ ED bisects BC.

ABCD is parallelogram.

Also given that BE = CE

In ABCD, AB || DC

∠DCE = ∠EBF … Alternate angles with transversal DF

In ∆DCE and ∆BFE, we have,
∠DCE = ∠EBF …from above

∠DEC = ∠BEF … Vertically opposite angles
Also, BE = CE … given
Hence, by ASA congruence rule,

∆DCE ≅​ ∆BFE
∴ DC = BF … by cpct

But DC = AB, as ABCD is a parallelogram.
∴ DC = AB = BF
Now, AF = AB + BF
From above, we get,
AF = AB + AB = 2AB
Hence, proved.

Here given that BC || QA and CA || QB which means that BCQA is a parallelogram.

∴ BC = QA …(1)

Similarly, BC || AR and AB || CR, which means BCRA is a parallelogram.

∴ BC = AR …(2)

But QR = QA + AR
From (1) and (2), we get,
QR = BC + BC

∴QR = 2BC

Hence, BC = QR

Here, Perimeter of ∆​ABC = AB + BC + CA

And Perimeter of ∆PQR =​PQ + QR + PR

Given that BC || QA and CA || QB which means BCQA is a parallelogram.

∴ BC = QA …(1)

Similarly, BC || AR and AB || CR, which means BCRA is a parallelogram.
∴ BC = AR …(2)

But, QR = QA + AR

From 1 and 2,

QR = BC + BC

∴QR = 2BC​

∴ BC = QR

Similarly, CA = ​ PQ and AB = ​ PR

Now,

Perimeter of ∆​ABC = AB + BC + CA

= QR + PQ + PR

= (PR + QR + PQ)

This states that,

Perimeter of ∆​ABC = (Perimeter of ∆​PQR)

∴ Perimeter of ∆PQR = 2 ⨯ Perimeter of ∆ABC

Here, ABCD is trapezium.

Join BD to cut EF at O.

It is given that, in ∆DAB, E is the mid point of AD and EO || AB.

∴O is the midpoint of BD …By converse of mid point theorem

Now in ∆BDC, O is the mid point of BD and OF || DC.
∴ F is the midpoint of BC … By converse of mid point theorem

Here, ABCD is parallelogram.

By the properties of parallelogram,

AD || BC and AB || DC

AD = BC and AB = DC

Also,

AB = AE + BE and DC = DF + FC

This means that,

AE = BE = DF = FC

Now, DF = AE and DF || AE, that is AEFD is a parallelogram​​.

Hence, AD || EF​

Similarly, ​BEFC is also a parallelogram.

Hence, EF || BC
∴​ AD || EF || BC

Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.

Also, the lines AD, EF and BC​ are also cut by the transversal AB at A, E and B, respectively such that​ AE = BE. ​
Similarly, they are also cut by​ GH.

Hence by intercept theorem,
∴ GP = PH

Hence proved.

Here, ABCD is trapezium.

Hence, AB|| DC

Also given that AP = PD and BQ = CQ

(i) In ∆QCD and ∆QBE, we have,
∠DQC =​∠BQE …Vertically opposite angles

∠DCQ = ∠EBQ …Alternate angles with transversal BC
BQ=CQ … P is the midpoint

Hence, by AAS test of congruency,
∆QCD ≅ ∆QBE
​Hence, DQ = QE …by cpct

(ii) Also, in ∆ADE, P and Q are the midpoints of AD and DE respectively

∴ PQ ||AE

Hence,PQ ||AB||DC

ie. AB || PR || DC

(iii) PQ, AB and DC are cut by transversal AD at P such that AP = PD.
Also they are cut by transversal BC at Q such that BQ = QC.
​ Similarly, lines PQ, AB and DC are also cut by AC at R.

Hence, by intercept theorem,
∴ AR = RC

In ∆ABC, AD is median.

∴ BD = DC

We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.

So, in ∆ABC, D is the mid point of BC and DE || BA.

Hence, DE bisects AC.

∴AE = EC

This means that E is the midpoint of AC.

∴BE is median of ∆ABC.

Here in AD and BE are medians.

Hence, in ∆ABC, we have:
AC = AE + EC

But AE = EC … as E is midpoint of AC

∴AC = 2EC …(1)

Nowin ∆BEC,

DF || BE

Also, EF = CF … by midpoint theorem, as D is the midpoint of BC

But,

EC = EF + CF

∴EC = 2 CF …(2)

From 1 and 2, we get,

AC = 4 CF

∴

ABCD is parallelogram.

(i) In ∆​ DCG, we have:

DG || EB
DE = EC … E is the midpoint of DC)
Also, GB = BC … by midpoint theorem
∴​ B is the midpoint of GC.
Also, GC = GB + BC
GC = 2BC

GC =​ 2 AD …as AD = BC

∴AD =GC

(ii) Now, in ∆​ DCG, DG || EB and E is the midpoint of DC and B is the midpoint of GC.

∴ EB = DG … by midpoint theorem

∴ DG = 2 EB

Let triangle be ∆ABC. D, E and F are the midpoints of sides AB, BC and CA, respectively.

By midpoint theorem, for D and E as midpoints of sides AB and BC,

DE∣∣AC

Similarly, DF∣∣BC and EF​∣∣AB.

∴ ADEF, BDFE and DFCE are all parallelograms.

But, DE is the diagonal of the BDFE.

∴ ​∆BDE≅​∆FED…(1)

Similarly, DF is the ​diagonal of the parallelogram ADEF.

∴ ∆DAF ≅ ∆FED …(2)
And, EF is the ​diagonal of the parallelogram DFCE.

∴ ∆EFC ≅ ∆FED …(3)

Hence, all the four triangles are congruent.

Here, in , are the midpoints of the sides and respectively.

By mid point theorem, as F and E are the mid points of sides AB and AC,

FE ∣∣ BC

Similarly, DE​∣∣ FB and FD​∣∣ AC.

Therefore, AFDE, BDEF and DCEF are all parallelograms.

We know that opposite angles in parallelogram are equal.

∴ In AFDE, we have,

∠A = ∠EDF

In BDEF, we have,

∠B = ∠DEF

In DCEF, we have,

∠ C = ∠​ DFE

Hence proved.

Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

Join diagonals of the rectangle.

In ∆ ABC, we have, by midpoint theorem,
∴ PQ ∣∣ AC and PQ = AC

Similarly, SR ∣∣ AC and SR =AC.

As, PQ ∣∣ AC and SR ∣∣ AC, then also PQ ∣∣ SR

Also, PQ = SR, each equal toAC …(1)

So, PQRS is a parallelogram

Now, in ∆SAP and ∆QBP, we have,

AS = BQ
∠A = ∠B = 90°
AP = BP

∴By SAS test of congruency,

∆SAP≅ ∆QBP​

Hence, PS = PQ …by cpct …(2)

Similarly, ∆SDR≅ ∆QCR

∴ SR = RQ … by cpct …(3)

Hence, from 1, 2 and 3 we have,

PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.

Hence, the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

In ΔABC, P and Q are mid points of AB and BC respectively.
∴ PQ|| AC and PQ = 1/2AC … (1) …Mid point theorem
Similarly in ΔACD, R and S are mid points of sides CD and AD respectively.
∴ SR||AC and SR = 1/2AC …(2) …Mid point theorem
From (1) and (2), we get
PQ||SR and PQ = SR
Hence, PQRS is parallelogram ( pair of opposite sides is parallel and equal)

Now, RS || AC and QR || BD.

Also, AC ⊥ BD … as diagonals of rhombus are perpendicular bisectors of each other.

∴RS ⊥ QR.

Thus, PQRS is a rectangle.

Hence, the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

Let ABCD be the square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

Join diagonals of the square.

In ∆ ABC, we have, by midpoint theorem,
∴ PQ ∣∣ AC and PQ = AC

Similarly, SR ∣∣ AC and SR =AC.

As, PQ ∣∣ AC and SR ∣∣ AC, then also PQ ∣∣ SR

Also, PQ = SR, each equal toAC …(1)

So, PQRS is a parallelogram

Now, in ∆SAP and ∆QBP, we have,

AS = BQ
∠A = ∠B = 90°
AP = BP

∴By SAS test of congruency,

∆SAP≅ ∆QBP​

Hence, PS = PQ …by cpct …(2)

Similarly, ∆SDR≅ ∆QCR

∴ SR = RQ … by cpct …(3)

Hence, from 1, 2 and 3 we have,

PQ = PQ = SR = RQ

We know that the diagonals of a square bisect each other at right angles.
∴ ∠​EOF = 90^o
​Now, RQ ∣∣ DB
⇒RE ∣∣ FO
Also, SR ∣∣ AC
⇒FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90^o​ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90^o and PQ = PS = SR = RQ.

This means that PQRS is square.

Hence, the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.

In ΔADC, S and R are the midpoints of AD and DC respectively.

By midpoint theorem,
Hence SR || AC and SR = AC … (1)
Similarly, in ΔABC, P and Q are midpoints of AB and BC respectively.
PQ || AC and PQ = AC …(2) …By midpoint theorem
From equations (1) and (2), we get
PQ || SR and PQ = SR …(3)
Here, one pair of opposite sides of quadrilateral PQRS is equal and parallel.
Hence PQRS is a parallelogram
Hence the diagonals of parallelogram PQRS bisect each other.
Thus PR and QS bisect each other.

Hence, the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

Here, in ABCD, diagonals intersect at 90°

Also, in ABCD, P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

In ∆ ABC, we have,
∴ PQ ∣∣ AC and PQ = AC …by midpoint theorem

Similarly, in ∆DAC,

SR ∣∣ AC and SR = AC …by midpoint theorem

Now, PQ ∣∣ AC and SR ∣∣ AC

∴PQ ∣∣ SR

Also, PQ = SR =AC

Hence, PQRS is parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.
∴ ∠​ EOF = 90°
​Also, RQ ∣∣ DB
∴ RE ∣∣ FO
Also, SR ∣∣ AC
∴ FR ∣∣ OE
∴ OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° …Opposite angles of parallelogram are equal
Thus, PQRS is a parallelogram with ∠R = 90^o.
∴​ PQRS is a rectangle.

Let the fourth angle be x

80^o + 95^o + 112^o + x^o = 360^o (Sum of angles of quadrilateral)

287^o + x^o = 360^o

x = 360^o – 287^o

= 73^o

Hence, option (B) is correct

Let the angles be 3x, 4x, 5x and 6x

3x + 4x + 5x + 6x = 360^o (Sum of angles of a quadrilateral)

18x = 360^o

x =

x = 20^o

∴ Angles of the quadrilateral are:

3x = 3 × 20^o = 60^o

4x = 4 × 20^o = 80^o

5x = 5 × 20^o = 100^o

6x = 6 × 20^o = 120^o

Hence, the smallest angle is 60^o

∴ Option (B) is correct

It is given in the question that,

In parallelogram ABCD: ∠ BAD = 75^o, ∠ CBD = 60^o

Now, ∠ DAB = ∠ DCB = 75^o (Opposite angles)

Also, in triangle DBC we know that sum of angles of a triangle is 180^o

∠ DBC + ∠ BDC + ∠ DCB = 180^o

60^o + ∠ BDC + 75^o = 180^o

135^o + ∠ BDC = 180^o

∠ BDC = 180^o – 135^o

∠ BDC = 45^o

Hence, option (C) is correct

As we know that from all the quadrilaterals given below, diagonals of a rectangle are equal

Hence, option (D) is correct

As we know that from all the quadrilaterals given below the diagonals of rhombus bisect each other at right angles

Hence, option (D) is correct

Let us assume a rhombus ABCD where,

AB = BC = CD = DA

Now, in triangle OBC by using Pythagoras theorem we get:

BC² = OB² + OC²

BC² = 6² + 8²

BC² = 36 + 64

BC² = 100

BC = √100

BC = 10 cm

∴ AB = BC = CD = DA = 10 cm

Hence, option (A) is correct

It is given in the question that,

ABCD is rhombus where, AB = BC = CD = DA

Now, by using Pythagoras theorem in triangle BOC we have:

BC² = OB² + OC²

(10)² = OB² + (8)²

100 = OB² + 64

OB² = 100 – 64

OB² = 36

OB = 6 cm

∴ Length of diagonal, BC = OB + OD

BC = 6 + 6

BC = 12 cm

Hence, option (B) is correct

It is given in the question that,

ABCD is a parallelogram where two adjacent angles ∠ A = ∠ B

We know that, sum of adjacent angles is 180^o

∴∠ A + ∠ B = 180^o

2∠ A = 180^o

∠ A = 180/2

∠ A = 90^o

As, ∠ A = ∠ B = ∠ C = ∠ D = 90^o

∴ ABCD is a rectangle as all the angles are equal to 90^o

Hence, option (C) is correct

It is given in the question that, ABCD is a quadrilateral where AO and BO are the bisectors of ∠ A and ∠ B

We know that, sum of all angles of a quadrilateral is equal to 360^o

∴∠ A + ∠ B + ∠ C + ∠ D = 360^o

∠ A + ∠ B + 70^o + 30^o = 360^o

∠ A + ∠ B = 360^o – 100^o

∠ A + ∠ B = 260^o

1/2 (∠A + ∠B) = 1/2 × 260°

1/2 (∠A + ∠B = 130°

Now, in triangle AOB

1/2 (∠A + ∠B) + ∠AOB = 180°

130^o + ∠ AOB = 180^o

∠ AOB = 180^o – 130^o

∠ AOB = 50^o

Hence, option (B) is correct

We know that,

Sum of two adjacent angles = 180^o

Also, sum of bisector of adjacent angles = 180/2 = 90^o

As sum of angles of a triangle = 180^o

∴ Sum of 2 adjacent angles + Intersection angle = 180^o

90^o + Intersection angle = 180^o

∴ Intersection angle = 180^o – 90^o

= 90^o

Hence, option (D) is correct

From all the given quadrilateral we know that the bisectors of the angles of a parallelogram enclose a rectangle

Hence, option (C) is correct

We know that, the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram

Hence, option (D) is correct

We know that, the figure formed by joining the mid-points of the adjacent sides of a square is a square

Hence, option (B) is correct

We know that, the figure formed by joining the mid-points of the adjacent sides of a parallelogram is parallelogram

Hence, option (D) is correct

We know that, the figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus

Hence, option (A) is correct

We know that, the figure formed by joining the mid-points of the adjacent sides of a rhombus is a rectangle

Hence, option (C) is correct

We know that,

Sum of two adjacent angles is equal to 180^o

∴∠ A + ∠ B = 180^o

According to the condition given in the question, we have

∠ A = x° then ∠ B = 2/3 x°

∴ x° + 2x/3 ° = 180°

5x/3 ° = 180°

⇒ x =

⇒ x = 540°/5

⇒ x = 108^o

∴∠ A = 108^o and,

∠ B = 2/3 × 108°

∠ B = 2× 36° = 72°

Thus, the smallest angle = ∠ B = 72^o

Hence, option (C) is correct

As per the question,

Let the smallest angle be x° and the largest angle be (2x – 24)°

Since, the sum of adjacent angles of a parallelogram is 180°

∴ x + (2x – 24) = 180°

3x – 24 = 180°

x = 68°

Hence, the largest angle is: 2x – 24 = 2(68) – 24 = 136 – 24 = 112

∴Option A is correct

As per the question,

∠ BAD = ∠ BCD = 75° (opposite angles of parallelogram)

Now, in ΔBCD,

∠BCD + ∠CBD + ∠BCD = 180°

45 + ∠CBD + 75 = 180°

∠CBD = 60°

∴ Option C is correct

Let the height of the parallelogram be ‘h’

Now, h < b (Since, perpendicular distance is the shortest)

∴ a × h < a × b

A < B

∴Option C is correct

According to the condition given in the question, we have

In triangle DCE and FBE

BE = EC (E is the mid-point of BC)

∠ CED = ∠ BEF (Vertically opposite angles)

∠ CDE = ∠ EFB (Alternate interior angles)

∴∆DCE∆FBE (By AAS congruence rule)

DC = BF (BY CPCT)

As AB is parallel to DC, then AB = DC

∴ AB = DC = BF

AF = AB + BF

AF = AB + AB

AF = 2AB

Hence, option (B) is correct

It is given in the question that,

ABCD is a trapezium

Draw EF parallel to AB and DC, and join BD intersecting EF at point M.

Now, E is the midpoint of AD and EM ∥ AB. Hence, using midpoint theorem,

EM = 1/2 AB

⇒ EM = 1/2 b

Similarly, FM = 1/2

⇒ DC = 1/2 a

EF = EM + FM

EF = 1/2 a + 1/2 b

EF = 1/2 (a + b)

∴Option B is correct

Construction: Join CF and extent it to cut AB at point M

Firstly, in triangle MFB and triangle DFC

DF = FB (As F is the mid-point of DB)

∠DFC = ∠MFB (Vertically opposite angle)

∠DFC = ∠FBM (Alternate interior angle)

∴ By ASA congruence rule

∆MFB ≅ DFC

Now, in triangle CAM

E and F are the mid-points of AC and CM respectively

∴ EF = 1/2 (AM)

EF = 1/2 (AB – MB)

EF = 1/2 (AB-CD)

Hence, option D is correct

Since, ABCD is a parallelogram,

∴∠B = ∠D (opposite angle)

1/2 ∠B = 1/2 ∠D

∠ADB = ∠ABD

∴ ADB is an isosceles triangle.

Since, M is the midpoint of BD

∴ AM is a median of ΔADB.

Now, ∠AMB = 90° (AM is perpendicular to BD)

∴Option C is correct

Since, we know that the diagonals of a rhombus bisect each other at 90°.

Hence, OA = AC, OB = BD and ∠​AOB ​= 90°

AB² = OA² + OB²

AB² = ( AC)² + ( BD)²

= (AC)² + (BD)²

AB² = (AC² + BD²)

4AB² = (AC² + BD²)

∴ Option C is correct

Draw perpendicular from D on AB meeting it on E and from C on AB meeting AB at F

∴​ DEFC will be a parallelogram and thus, EF = CD

Now, In ∆ABC

Since, ∠B is acute

∴ AC² = BC² + AB² - 2AB × AE (i)

Similarly, In ∆ABD,

Since ∠A is acute​

∴ ​BD² = AD² + AB² - 2AB × AF (ii)

Adding (i) and (ii),

AC² + BD² = (BC² + AD²) + (AB² + AB²) - 2AB (AE + BF)

= (BC² + AD²) + 2AB (AB - AE - BF) [Since, AB = AE + EF + FB and AB - AE = BE]

= (BC² + AD²) + 2AB (BE - BF)

= (BC² + AD²) + 2AB.EF

Now, we know that CD = EF

Thus, AC² + BD² = ​(BC² + AD²) + 2AB.CD

∴ Option D is correct

We know that,

Area of a parallelogram = base ⨯ height

Now, if both parallelograms are on the same base and between the same parallels, then their heights will be equal.

Hence, their areas will also be equal

∴ Option D is correct

Let G be the mid-point of FC and join DG

​In ∆BCF,

G is the mid-point of FC and D is the mid-point of BC

Thus, DG|| BF

DG || EF

Now, In ∆ ADG,

E is the mid-point of AD and EF is parallel to DG.

Thus, F is the mid-point of AG.

AF = FG = GC [G is the mid-point of FC]

Hence, AF = AC​

∴ Option B is correct

Let the required angles be 3x, 7x, 6x and 4x

3x + 7x + 6x + 4x = 360° (Sum of angles of quadrilateral)

20x = 360°

x = 18°

​Hence, angles are:

3x = 3 ⨯18° = 54°

7x = 7 ⨯ 18° = 126°

6x = 6 ⨯ 18° = 108°

4x = ​4 ⨯18° = 72°

Now we can observe that, 54° + 126° = 180° and 72° + 108° = 180°

Thus, ABCD is a trapezium.

Hence option C is correct.

We know that,

In any parallelogram, opposite angles are bisected by the diagonals

∴ Option C is correct

It is given in the question that,

APB and CQD are two parallel lines,

Thus, the bisectors of ∠CQP, ∠APQ, ∠BPQ and ∠PQD enclose a rectangle.

Hence, option C is correct.

In the given figure,

∠OAD = ​∠OCB (Alternate interior angle)

​∠OCB = 30°

∠AOB + ∠BOC = 180° (Linear pair)

70° + ∠BOC = 180°

∠BOC = 110°

Now, In ∆BOC,

∠OBC + ∠BOC + ∠OCB = 180°

∠OBC + 110° + 30° = 180°

∠OBC = 40°

∴ ​∠DBC = 40°

Hence, Option A is correct.

We can clearly observe that statement I and statement II are correct. Whereas Statement III is not correct because the triangle formed by joining the midpoints of the sides of an isosceles triangle is always an isosceles triangle

Therefore, Option C is correct

We can clearly observe that statement II and statement III are correct and Statement I is wrong because the diagonals of a rectangle does not bisect ∠A and ∠C. And this is so because the adjacent sides are unequal in a rectangle.

∴ Option B is correct

Here, as we know that if the diagonals of a ​quadrilateral bisects each other, then it is a parallelogram.

But as per II, if the diagonals of a quadrilateral are equal, then it is not necessarily a ​parallelogram which is not true. Thus, II does not give the answer.

Therefore Option A is correct.

Here, we can observe that neither I not II can alone justify the answer to the given question. But if we consider both I and II together then they completely satisfies the answer.

∴ Option C is correct.

We know that when the diagonals of a parallelogram are equal, it might be a square or a rectangle. But if the diagonals of that parallelogram intersect at a right angle, then it is definitely a square. Thus, it can be concluded that both I and II together will give the answer.

Therefore, Option C is correct.

We know that a quadrilateral is a parallelogram when either I or II holds true.

​Hence, the correct answer is (b)

Let the fourth angle be x,

130^o + 70^o + 60° + x° = 360° (angle sum of quadrilateral)

x° = 360^o − (130^o + 70^o + 60^o)

x° = 100^o

Thus, it can be observed that reason and assertion both are true and the reason explains the assertion.

Therefore Option A is correct.

It is given that, ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then, PQRS is a parallelogram

Also, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Hence, both assertion and reason are true and reason is correct explanation of the assertion

∴ Option (a) is correct

It is given that,

In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠ C which is true

And we know that, the diagonals of a rhombus bisect each other at right angles.

Hence, both assertion and reason are true but reason is not the correct explanation of assertion

∴ Option (b) is correct

The statement given in assertion is not true as every parallelogram is not a rectangle whereas, statement given in the reason is true as the angle bisectors of a parallelogram form a rectangle

Hence, assertion is false whereas reason is true

∴ Option (d) is correct

We know that,

The diagonals of a ||gm bisect each other

Also we know that, if the diagonals of a ||gm are equal and intersect at right angles, then the parallelogram is a square

Hence, both assertion and reason are true but reason is not the correct explanation of the assertion

Hence, option (b) is correct

The correct match for the above given table is as follows:

a) PQ = (AB + CD)

PQ = (17)

PQ = 8.5 cm

(b) OR = (PR)

OR = ​ (13)

OR = 6.5 cm

from the above given four statements option A is false as we know that in any parallelogram the diagonals are not equal

Hence, option A is correct

In ∆ABC,

Since, AD is the median

Thus, BD = DC

Let the height of ∆ABC be h

ar(∆ABD) = ar(∆ABD)

1/2 × h × BD = 1/2 × h × BD

1/2 × h × BD = 1/2 × h × CD

∴ ar (∆ABD) = ar (∆ADC)

Let H be the height of ∆BPD and ∆PDC

∴ ar (∆BPD) = ar (∆PDC)​

Now, ar(∆ABD) = ar (∆ABP) + ar (∆BPD)

And, ar(∆ACD) = ar(∆ACP) + ar(∆PDC)

Thus, ar(∆ABP) = ar(∆ACP)​

∴ Option A is correct

Let the angles be x, 3x, 5x and 6x.

x + 3x + 5x + 6x = 360° (sum of angles of quadrilateral)

15x° = 360^o

x° = 24^o

Therefore, angles are as follows:

x° = 24^o

3x° = 24^o⨯ 3 = 72^o

5x° = 24^o⨯ 5 = 120^o

6x° = 24^o⨯ 6 = 144^o

Hence, 144° is the greatest angle.

We know that in ∆ABC, D and E are the midpoints of AB and AC, respectively.

Now using mid-point theorem,

DE = (BC)

BC= 2 ⨯ DE

BC= 2 ​⨯ 5.6

= 11.2 cm

Thus, BC = 11.2 cm

In ∆ABC, using mid point theorem

We know that D is the mid-point of BC and DE|| AB.

Thus, AE = EC and DE = (AB)

Now, E is the mid point of AC

Thus, BE is the median

Here, we have:

l ||​ m ||​ n

And p and q are the transversal lines

Thus, AB : BC = 5 : 15

AB : BC = 1 : 3

∴ Using intercept theorem,

DE : EF = 1 : 3

Let there be a rectangle ABCD with AB = CD and BC = AD and ∠A = ∠B = ∠C= ∠D = 90^o

Since, BD bisects ∠B

∠​ABD = ∠​DBC (i)

​And, ∠​ADB = ​∠​DBC [Alternate interior angles]

​∠​ABD = ∠​ADB. [From (i)]

AB = DA. (Sides opposite to equal angles)

∴ AB = CD = DA = BC

Since, all the sides are equal and all the angles are equal to 90^o, thus the quadrilateral is a square.

Hence, ABCD is a square.

∠BOC = ​∠AOD (Vertically opposite angles)

Angle AOD = 50°

In ∆ AOD, Since, the diagonals are equal, thus the bisectors will also be equal)

Thus, OA = OD

∴ ​∠OAD = ​∠ODA

= (180° - 50°)

= (130°)

= 65°

∴ Option C is correct

The correct match for the above given table is as follows:

∠BDC = ∠ABD (Alternate interior angles)

∠ABD = 50^o

Now, In ∆AOB,

∠DBA = 50^o and ∠AOB = 90^o

Thus, ∠OAB = 180^o − (90^o + 50^o)

∠OAB = 180° - 140°

∠OAB = 40^o

∴ Option B is correct.

Construction: Draw perpendicular line from D and C to AB such that it cuts AB at F and E, respectively.

Now, In ∆​ADF and ∆BCE,

AD = BC (Given)

∠AFD = ∠BEC (90^o each)

DF = CE (Perpendicular distance between the same parallels)

∴ By SSA axiom

∆​ADF ≅ ∆BCE​

∠A​ = ∠B (by c.p.c.t.)

Therefore Option A is correct.

We can clearly observe that statement I and statement III are correct.

We can prove the statement as follows:

In ​∆ABC, altitudes AD, BE and CF are equal

Now, In ∆ABE and ∆ACF,

BE = CF (Given)

∠A = ∠A (common)

∠​AEB = ∠​AFC (Each 90°)

Therefore, by AAS axiom,

∆ABE ≅​ ∆ ACF

AB = AC (by cpct)

In the same way, ∆BCF ≅​ ∆BAD

thus, BC = AB (by cpct)

Therefore AB = AC = BC

Thus, ∆ABC is an equilateral triangle.

We can prove the IIIrd statement as follows:

Let ​∆ABC be an isosceles triangle with AD as an altitude

Now, In ​ ​∆ABD and ​∆ADC,

AB = AC (Given)

∠B = ​​​∠C (Angles opposite to equal sides)

∠BDA = ​∠CDA (each 90°)

Therefore by AAS axiom,

∆ABD ≅ ​​∆ADC

BD = DC (by congruent parts of congruent triangles)

∴ D is the mid-point of BC and hence AD bisects BC.

Area of a triangle = 1/2 (Base × Height)

Now, draw AL perpendicular to BC and h be the height of ∆ABC i.e. AL

Thus, Height of ∆ABD ​= Height of ∆ADE ​ = Height of ∆AEC

It is given that the bases BD, DE and EC of ∆ABD, ∆ADE and ∆AEC respectively are equal.

Now, since base and height both are equal of all the triangles therefore,

ar(∆ABD) = ar(∆ADE) = ar(∆AEC)

Now, here in ∆ADE and ∆BCF,

AD = BC (Opposite sides of parallelogram ABCD

DE = CF (Opposite sides of parallelogram DCEF)

AE = BF ​(Opposite sides of parallelogram ABFE)​

​​∴ By SSS axiom,

∆ADE ≅​ ∆BCF

And,

ar(∆ADE) = ar(∆BCF) (By cpct)

Here, in trapezium ABCD,

AB || DC and AC and BD are the diagonals intersecting at O.

Now, since ​∆ACD and ∆BCD lie on the same base and between the same parallels.

Thus, ar(∆ACD) = ar(∆BCD) ​

Subtracting ar(∆COD) from both the sides, we get:

ar(∆ACD) − ar(∆COD) = ar(∆BCD)− ar(∆COD)​

∴ ar(∆AOD) = ar(∆BOC)

Let there be a parallelogram ABCD and with one of its diagonal as AC.

Now, In ​∆CDA and ∆ABC,

DA = BC (Opposite sides of parallelogram ABCD)

AC = AC (Common)

CD = AB (Opposite sides of parallelogram ABCD)

∴ By SSS axiom

∆CDA ≅ ∆ABC

ar(∆CDA) = ar(∆ABC) ​ (by cpct)

Thus, we can say that the diagonal of a parallelogram divides it into two triangles of equal area.

Here we have ABCD as a quadrilateral with one of its diagonal as AC and BL and DM are perpendicular to AC

Thus, ar(ABCD) = ar(∆ADC) + ar(∆ABC)

Since, (BL ⊥ AC) and (DM ⊥ AC)

∴ Area of ABCD = ( ⨯​ AC ⨯​ BL) + ( ⨯​ AC ⨯​ DM)

= ⨯​ AC ⨯​ (BL + DM)

Here we know that parallelogram ABCD and rectangle ABEF are on the same base AB and between the same parallels such that:

AB = CD and AB = EF

So, CD = FE

Now, adding AB on both sides

AB + CD = AB + FE (i)

Since we know that hypotenuse is the longest side of a triangle

∴ AD > AF (ii)

And, BC > BE (iii)

Adding (ii) and (iii),

​ AD + BC > AF + BE (iv)

Now, Perimeter of ABCD = AB + BC + CD + AD

And, Perimeter of ABEF = AB + BE + FE + AF

​Adding (i) and (iv),

AB + CD + AD + BC > AB + FE + AF + BE

Thus, we can say that the perimeter of parallelogram ABCD is greater than that of rectangle ABEF.

Here we have parallelogram ABCD with AB ∣∣ DC

Thus, DC ∣∣​ BF

Now, in ∆DEC and ∆FEB,

∠DCF = ∠EBF (Alternate interior angle)

CE = BE (E is the mid-point of BC

∠CED = ∠BEF (Vertically opposite angle)

Therefore, by ASA axiom,

∆​DEC ≅ ∆FEB

CD = BF (by cpct)

And CD = AB (Opposite sides of a parallelogram ABCD)

So, AF = AB + BF = AB + AB = 2AB ​

(i) Here, we have

∠CRQ = ∠CBA = 90^o

Thus, RQ ∣∣​ AB

Now, In ∆ABC,

Q is the mid-point of AC and QR ∣∣​ AB​.

Thus, R is the mid-point of BC.

In the same way, P is the midpoint of DC.

Hence, DP = PC

(ii) Here, let us join B to D.

Now, In ∆​CDB,

P and R are the mid points of DC and BC respectively.

Since, AC = BD

Thus, PR ∣∣​ DB and PR = DB = AC