Quadrilaterals

Given: 3 angles of quadrilateral - 75°, 90° and 75°

Let the fourth angle be x.

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ 75° + 90° + 75° + x = 360°

⇒ 240° + x = 360°

⇒ x = 360° – 240°

⇒ x = 120°

Hence, the fourth angle is 120°.

Given: Angle between a side of rectangle and its diagonal = 25°

Let the acute angle between diagonals be x

Now, we know that diagonals of a rectangle are equal in length i.e.,

AC = BD

Dividing both sides by 2,

⇒ 1/2AC = 1/2BD

⇒ OD = OC [∵ O is mid-point of AC and BD]

⇒ ∠y = 25° [∵ angles opposite to equal sides are equal]

Now we know that, exterior angle is equal to the sum of two opposite interior angles.

∴ ∠BOC = ∠ODC + ∠OCD

⇒ ∠x = ∠y + 25°

⇒ ∠x = 25° + 25°

⇒ ∠x = 50°

Hence, the acute angle between diagonals is 50°.

Given: ABCD is a rhombus

∠ACB = 40°

∵ ∠ACB = 40°

⇒ ∠OCB = 40°

∵ AD ∥ BC

⇒ ∠DAC = ∠BCA = 40° [Alternate interior angles]

⇒ ∠DAO = 40°

Since, diagonals of a rhombus are perpendicular to each other

∴ ∠AOD = 90°

Sum of all angles of a triangle is 180°

⇒ ∠AOD + ∠ADO + ∠DAO = 180°

⇒ 90° + ∠ADO + 40° = 180°

⇒ 130° + ∠ADO = 180°

⇒ ∠ADO = 180° – 130°

⇒ ∠ADO = 150°

⇒ ∠ADB = 150°

Hence, ∠ADB = 150°

Since, diagonals of rectangle are equal

∴ AC = BD

⇒ PQ = QR

∴ PQRS is a rhombus

Diagonals of a rhombus are perpendicular.

Hence, diagonals of PQRS are perpendicular

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

∵ ABCD is a rhombus

∴ AB = BC = CD = DA

Now,

∵ D and C are midpoints of PQ and PS

∴ DC = 1/2QS [By midpoint theorem]

Also,

∵ B and C are midpoints of SR and PS

∴ BC = 1/2PR [By midpoint theorem]

∵ ABCD is a rhombus

∴ BC = CD

⇒ 1/2QS = 1/2PR

⇒ QS = PR

Hence, diagonals of PQRS are equal

Given: Ratio of angles is 3: 7: 6: 4

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

Sum of all angles of a quadrilateral is 360°

⇒ 3x + 7x + 6x + 4x = 360°

⇒ 20x = 360°

⇒ x = 18°

Now,

Angles of quadrilateral are

∠A = 3x = 3 × 18° = 54°

∠B = 7x = 7 × 18° = 126°

∠C = 6x = 6 × 18° = 108°

∠D = 4x = 4 × 18° = 72°

Extend DC to E

Now,

∠BCE + ∠BCD = 180° [∵ sum of angles on a straight line = 180°]

⇒ ∠BCE + ∠BCD = 180°

⇒ ∠BCE + 108° = 180°

⇒ ∠BCE = 180° – 108°

⇒ ∠BCE = 72°

∵ ∠BCE = ∠ADC = 72°

∴ BC ∥ AD

Hence, it is a trapezium.

Sum of all angles of a quadrilateral is 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

On dividing both sides by 2,

⇒ 1/2(∠A + ∠B + ∠C + ∠D) = 1/2 × 360° = 180°

∵ AP, PB, RC and RD are bisectors of ∠A, ∠B, ∠C and ∠D

⇒ ∠PAB + ∠ABB + ∠RCD + ∠RDC = 180° …(1)

Sum of all angles of a triangle is 180°

∴ ∠PAB + ∠APB + ∠ABP = 180°

⇒ ∠PAB + ∠ABP = 180° – ∠APB …(2)

Similarly,

∴ ∠RDC + ∠RCD + ∠CRD = 180°

⇒ ∠RDC + ∠RCD = 180° – ∠CRD …(3)

Putting (2) and (3) in (1),

180° – ∠APB + 180° – ∠CRD = 180°

⇒ 360° – ∠APB – ∠CRD = 180°

⇒ ∠APB + ∠CRD = 360° – 180°

⇒ ∠APB + ∠CRD = 180° …(4)

Now,

∠SPQ = ∠APB [vertically opposite angles]

∠SRQ = ∠DRC [vertically opposite angles]

Putting in (4),

⇒ ∠SPQ + ∠SRQ = 180°

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Let the bisectors of the angles APQ and CQP meet at M and bisectors of the angles BPQ and PQD meet at N.

Join PM, MQ, QN and NP.

∠APQ = ∠PQD [∵APB ∥ CQD]

⇒ 2∠MPQ = 2∠NQP [∵ NP and PQ are angle bisectors]

Dividing both sides by 2,

⇒ ∠MPQ = ∠NQP

⇒ PM ∥ QN

Similarly,

⇒ ∠BPQ = ∠CQP

⇒ PN ∥ QM

∴ PNQM is a parallelogram

Now,

∠CQP + ∠CQP = 180° [Angles on a straight line]

⇒ 2∠MQP + 2∠NQP = 180°

Dividing both sides by 2,

⇒ ∠MQP + ∠NQP = 90°

⇒ ∠MQN = 90°

Hence, PMQN is a rectangle

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

Join AC, RP and SQ

In ∆ABC,

P is midpoint of AB and Q is midpoint of BC

∴ By midpoint theorem,

PQ ∥ AC and PQ = 1/2AC …(1)

Similarly,

In ∆DAC,

S is midpoint of AD and R is midpoint of CD

∴ By midpoint theorem,

SR ∥ AC and SR = 1/2AC …(2)

From (1) and (2),

PQ ∥ SR and PQ = SR

⇒ PQRS is a parallelogram

ABQS is a parallelogram

⇒ AB = SQ

PBCR is a parallelogram

⇒ BC = PR

⇒ AB = PR [∵ BC = AB, sides of rhombus]

⇒ SQ = PR

∴ diagonals of the parallelogram are equal

Hence, it is a rectangle.

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

In ∆ABC,

D and E are midpoints of AB and AC

By midpoint theorem,

DE ∥ BC and DE = 1/2BC

⇒ DE = 1/2[BP + PO + OQ + QC]

∵ P and Q are midpoints of OB and OC

⇒ DE = 1/2[2PO + 2OQ]

⇒ DE = PO + OQ

⇒ DE = PQ …(1)

Now,

In ∆AOC,

Q and C are midpoints of OC and AC

By midpoint theorem,

EQ ∥ AO and EQ = 1/2AO …(2)

Similarly,

In ∆AOB,

D and P are midpoints of AB and OB

By midpoint theorem,

PD ∥ AO and PD = 1/2AO …(3)

Also,

By midpoint theorem,

DE ∥ BC and DE = 1/2BC …(4)

From (2) and (3),

EQ ∥ PD and EQ = PD

From (1) and (4),

DE ∥ BC

DE ∥ PQ and DE = PQ

Hence, DEPQ is a parallelogram

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

PQRS is a square

⇒ PQ = QR = RS = SP

Also, PR = SQ [Diagonals of a square]

And,

BC = PR

SQ = AB

⇒ BC = PR = SQ = AB

i.e, BC = AB

Thus all sides of ABCD are equal

⇒ ABCD is a square or a rhombus.

In ∆ABD,

S and P are midpoints of AD and AB

By midpoint theorem,

SP ∥ DB and SP = 1/2DB

Similarly,

In ∆ABC,

Q and P are midpoints of BC and AB

By midpoint theorem,

PQ ∥ AC and PQ = 1/2AC

∵ PQRS is a square

SP = PQ

⇒ 1/2DB = 1/2AC

⇒ DB = AC

i.e, Diagonals of ABCD are equal

∴ it cannot be a square

Hence, it is a rhombus

Given: ∠AOB = 70°

∠DAC = 32°

∵ AD ∥ BC and AC is transversal

∴ ∠ACB = 32°

Now,

∠AOB + ∠BOC = 180°

⇒ 70° + ∠BOC = 180°

⇒ ∠BOC = 180° - 70°

⇒ ∠BOC = 110°

Sum of all angles of a triangle = 180°

⇒ ∠BOC + ∠BCO + ∠OBC = 180°

⇒ 110° + 32°+ ∠OBC = 180°

⇒ 142°+ ∠OBC = 180°

⇒ ∠OBC = 180° - 142°

⇒ ∠OBC = 38°

Hence, ∠DBC = 38°

In a parallelogram the opposite angles are not bisected by the diagonals. This statement is false.

But,

In a parallelogram,

Opposite sides are equal

Opposite angles are equal

Diagonals bisect each other

All these statements are true.

Hence, opposite angles are bisected by the diagonals is not true for a parallelogram.

Let us assume that, DE = EF

AE = CE [∵ E is mid-point of AC]

DE = EF [assumed]

∠AED = ∠FEC [vertically opposite angles]

∴ By SAS, ∆AED ≅ ∆FEC

∴ By CPCT, AD = CF and ∠ADE = ∠CFE

∵ alternate interior angles are equal

∴ AD ∥ CF

That proves our assumption was correct

Hence, DE = EF

Given: OA = 3 cm

OD = 2 cm

We know that,

Diagonals of parallelogram bisect each other.

∴ AC = 2AO

= 2 × 3 cm

= 6 cm

And,

BD = 2OD

= 2 × 2 cm

= 4 cm

Hence, AC = 6 cm and BD = 4cm

Diagonals of a parallelogram bisect each other but not at 90°.

So, they are not perpendicular to each other.

Hence, this statement is false.

We know that,

Sum of all angles of a quadrilateral = 360°

But here,

110° + 80° + 70° + 95° = 355° ≠ 360°

Hence, 110°, 80°, 70° and 95° cannot be the angles of a quadrilateral.

We know that,

Sum of co-interior angles is 180° in a trapezium

Hence, it is a trapezium.

Let all the angles be equal to x

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ x + x + x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

Hence, the quadrilateral is a rectangle.

Diagonals of a rectangle bisect each other therefore they are equal but they are not perpendicular.

Hence, the statement is not true.

We know that, sum of all angles of a quadrilateral = 360°

So atleast one angle should be acute angle.

Hence, all the four angles of a quadrilateral cannot be obtuse angles.

Given: AB = 5 cm

BC = 8 cm

CA = 7 cm

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

By the midpoint theorem,

DE = �AC

⇒ DE = � × 7 cm

⇒ DE = 3.5 cm

Hence, the length of DE = 3.5 cm

∵ BDEF is a parallelogram

∴ BD = FE and BD ∥ FE …(1)

Similarly,

∵ FDCE is a parallelogram

∴ CD = FE and CD ∥ FE …(2)

From (1) and (2),

BD = CD and BD ∥ CD

Hence, BD = CD.

Given: C = 55°

Opposite angles of a parallelogram are equal

∵ ABCD is a parallelogram

∴ ∠A = ∠C = 55°

Also,

∵ AEFG is a parallelogram

∴ ∠F = ∠A = 55°

Hence, ∠F = 55°.

We know that, sum of all angles of a quadrilateral = 360°

So atleast one angle should be obtuse angle.

Hence, all the four angles of a quadrilateral cannot be acute angles.

We know that, sum of all angles of a quadrilateral = 360°

If all angles are right angle then,

90° + 90° + 90° + 90° = 360°

Hence, all the angles of a quadrilateral can be right angles.

Given: A = 35°

∵ diagonals of the quadrilateral ABCD bisect each other

∴ it is a parallelogram

⇒ A + B = 180°

⇒ 35° + B = 180°

⇒ B = 180° – 35°

⇒ B = 145°

Hence, B = 145°.

Given: AB = 4 cm

Since, opposite angles of the quadrilateral are equal, therefore, it is a parallelogram.

Hence, its opposite sides are also equal.

⇒ CD = AB = 4 cm

Hence, CD = 4 cm

Let the remaining three equal angles be x.

We know,

Sum of all interior angles of a quadilateral is

Since it is an isosceles trapezium.

We know,

Angles opposite to each other in quadrilateral are supplementary.

Similarly,

Given ABCD is parallelogram in which DP is perpendicular to AB and DQ is perpendicular to BC.

Given,

In quad. DPBQ, by angle sum property we have,

So (opposite angles in parallelogram are equal)

Since, AB||CD (opposite sides are parallel in parallelogram),

(Sum of adjacent interior angles is )

So

Given,

ABCD is a rhombus.DE is the altitude on AB then AE = EB.

In a ΔAED and ΔBED,
DE = DE (common line)
∠AED = ∠BED (right angle)
AE = EB (DE is an altitude)
∴ ΔAED ΔBED (SAS property)

∴ AD = BD (by C.P.C.T)

But AD = AB (sides of rhombus are equal)
⇒ AD = AB = BD
∴ ABD is an equilateral triangle.
∴ ∠A = 60
⇒ ∠A = ∠C = 60 (opposite angles of rhombus are equal)
But Sum of adjacent angles of a rhombus is supplementary.
∠ABC + ∠BCD = 180
∠ABC + 60 = 180
∠ABC = 180-60 = 120
∴ ∠ABC = ∠ADC = 120 (opposite angles of rhombus are equal)
∴ Angles of rhombus are ∠A = 60, ∠C = 60, ∠B = 120, ∠D = 120

Join BD, meeting AC at O.

Since diagonals of a parallelogram bisect each other,

OA = OC and OD = OB.

So,

OA = OC and AE = CF,

OE = OF

Now, BFDE is a quadrilateral whose diagonals bisect each other.

Hence, BFDE is a parallelogram.

The figure for the solution:

ABCD is a trapezium in which AB||DC.

Joining diagonal AC, which intersects EF at O.

In ΔADC, E is mid-point of AD and OE || CD.

By mid-point theorem, O is mid-point of AC.

Similarly,

In ΔCBA, O is mid-point of AC and OF||AB.

F is mid-point of BC by Mid-point Theorem.

Given,

PQ||AB, PR||AC and RQ||BC.

In quadrilateral BCAR,

BR||CA and BC||RA

BCAR is a parallelogram

BC = AR …(i)

Now, in quadrilateral BCQA,

BC||AQ and AB||QC

BCQA is a parallelogram

BC = AQ …(ii)

Adding Eqn. (i) and (ii), we get

2BC = AR + AQ

2BC = RQ

Hence, proved.

Given, D, E and F are mid-points od sides BC, CA and AB, respectively.

So, EF||BC and by Mid-Point Theorem,

DF||AC and DE||AB

Also, EF = BC, DE = AB and FD = AC

Since, ABC is an equilateral triangle,

AB = BC = AC

DE = EF = FD

Thus, all sides of ΔDEF are equal.

Hence, ΔDEF is an equilateral triangle.

Joining AQ and PC.

Since ABCD is a parallelogram, AB||DC and AP||QC.

Given, AP = CQ.

APCQ is a parallelogram.

We know, diagonals of a parallelogram bisect each other.

Hence AC and PQ bisect each other.

Since ABCD is a parallelogram, AD||BC and AB is a transversal.

(Sum of co-interior angles is 180)

In ΔABP,

(opposite sides of equal angles are equal)

Now, multiply both sides by 2, we get,

(P is mid-point of BC)

(AB||CD and AD||BC)

Hence, Proved.

Given, ΔABC with ∠A = 90 and

AB = AC …(i)

Let ADEF is a square,

AD = AF = EF = AD …(ii)

Subtracting eqn (ii) from (i),

AB-AD = AC-AF

BD = CF

Now, in ΔCFE and ΔEDB,

BD = CF

DE = EF

∠CFE = ∠EDB = 90 (Side of square)

ΔCEF BED (By SAS)

CE = BE

Hence, vertex E of the square bisect the hypotenuse BC.

Given, AB = 10 cm, AD = 6 cm
DC = AB = 10 cm and AD = BC = 6 cm
Given, bisector of ∠A intersects DE at E and BC produced at F.
Now, drawing PF || CD.
From the figure, CD || FP and CF || DP
PDCF is a parallelogram.
And , AB || FP and AP || BF
ABFP is also a parallelogram
In ΔAPF and ΔABF
∠APF = ∠ABF (opposite angles of a parallelogram are equal)
AF = AF (Common side)
∠PAF = ∠AFB (Alternate angles)
ΔAPF ≅ ΔABF (By ASA congruence criterion)
AB = AP (CPCT)
AB = AD + DP
= AD + CF (Since DCFP is a parallelogram)
∴ CF = AB – AD
= (10 – 6) cm = 4 cm

Given, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD.

In ΔADC, by mid-point theorem,

SR||AC and SR = AC

In ΔABC, by mid-point theorem,

PQ||AC and PQ = AC

SR = PQ = AC

Similarly,

In ΔBCD, by mid-point theorem,

RQ||BD and RQ = BD

In ΔBAD, by mid-point theorem,

SP||BD and SP = BD

SP = RQ = BD = AC

So,

SR = PQ = SP = RQ

Hence, PQRS is a rhombus.

Given, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also,

AC is perpendicular to BD

∠COD = ∠AOD = ∠AOB = ∠COB = 90

In ΔADC, by mid-point theorem,

SR||AC and SR = AC

In ΔABC, by mid-point theorem,

PQ||AC and PQ = AC

PQ||SR and SR = PQ = AC

Similarly,

SP||RQ and SP = RQ = BD

Now, in quad EOFR,

OE||FR, OF||ER

∠EOF = ∠ERF = 90

Hence, PQRS is a rectangle.

Given, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Also, AC = BD and AC is perpendicular to BD.

In ΔADC, by mid-point theorem,

SR||AC and SR = AC

In ΔABC, by mid-point theorem,

PQ||AC and PQ = AC

PO||SR and PQ = SR = AC

Now, in ΔABD, by mid-point theorem,

SP||BD and SP = BD = AC

In ΔBCD, by mid-point theorem,

RQ||BD and RQ = BD = AC

SP = RQ = AC

PQ = SR = SP = RQ

Thus, all four sides are equal.

Now, in quadrilateral EOFR,

OE||FR, OF||ER

∠EOF = ∠ERF = 90(Opposite angles of parallelogram)

∠QRS = 90

Hence, PQRS is a square.

Let ABCD is a parallelogram and diagonal AC bisect ∠A.

∠CAB = ∠CAD

Now,

AB||CD and AC is a transversal.

∠CAB = ∠ACD

Again, AD||BC and AC is a transversal.

∠DAC = ∠ACB

Now,

∠A = ∠C

∠A = ∠C

∠DAC = ∠DCA

AD = CD

But, AB = CD and AD = BC (Opposite sides of parallelograms)

AB = BC = CD = AD

Thus, ABCD is a rhombus.

Given, P and Q are mid-points of AB and CD.

Now, AB||CD,

AP||QC

Also, AB = DC

AB = DC

AP = QC

Now, AP||QC and AP = QC

APCQ is a parallelogram.

AQ||PC or SQ||PR

Again,

AB||DC AB = DC

BP = QD

Now, BP||QD and BP = QD

BPDQ is a parallelogram

So, PD||BQ or PS||QR

Thus, SQ||RP and PS||QR

PQRS is a parallelogram.

Given, ABCD is a quadrilateral in which AB || DC and AD = BC.

Extend AB to E and draw a line CE parallel to AD.

Since AD||CE and transversal AE cuts them at A and E, respectively.

∠A + ∠E = 180

∠A = 180-∠E

Since, AB||CD and AD||CE

So quadrilateral AECD is a parallelogram.

Now, AD = CE BC = CE

In ΔBCE,

CE = BC

∠CBE = ∠CEB (opposite angles of equal sides are equal)

180-∠B = ∠E

180-∠E = ∠B

∠A = ∠B

Hence, proved.

In quad ABED, we have AB||DE and AB = DE

ABED is a parallelogram.

AD||BE and AD = BE

Now, in quad ACFD, we have

AC||FD and AC = FD

ACFD is a parallelogram.

AD||CF and AD = CF

AD = BE = CF and CF||BE

Now, in quad BCFE,

BE = CF and BE||CF.

BCFE is a parallelogram.

BC = EF and BC||EF

Hence, proved.

Given, a ABC in which AD is a median and E is mid-point of AD. Now, drawing DP||EF.

In ΔADP, E is mid-point of AD and EF||DP.

F is mid-point of AP.

In ΔFBC, D is mid-point of BC and DP||BF.

P is mid-point of FC.

Thus, AF = FP = PC

Hence, AF = AC

Hence, proved.

Let ABCD is a square.

AB = BC = CD = AD

P,Q,R and S are mid-points of AB,BC,CD and DA, respectively.

Now, in ΔADC,

SR||AC and SR = AC

In ΔABC,

PQ||AC and PQ = AC

SR||PQ and SR = PQ = AC

Similarly,

SP||BD and BD||RQ

SP||RQ and SP = BD

And RQ = BD

SP = RQ = BD

Since, diagonals of a square bisect each other at right angles.

AC = BD

SP = RQ = AC

SR = PQ = SP = RQ

All sides are equal.

Now, in quad OERF,

OE||FR and OF||ER

∠EOF = ∠ERF = 90

Hence, PQRS is a square.

Given, ABCD is a trapezium in which AB||CD. Also, E and F are mid-points of sides AD and BC.

Now, joining BE and produce it to meet CD produced at G.

In ΔGCB, by mid-point theorem

EF||GC

But GC = CD, and CD||AB

EF||AB

Now, draw BD which intersects EF at O.

In ΔADB, AB||EO and E is mid-point of AD.

By converse of mid-point theorem, O is mid-point of BD.

Also, EO = AB …(i)

In ΔBDC, OF||CD and O is mid-point of BD.

OF = CD …(ii)

Adding (i) and (ii),

EO + OF = AB + CD

EF = (AB + CD)

Hence, proved.

Let ABCD is a parallelogram.

Since, DC||AB and DA is transversal.

∠A + ∠D = 180

∠A + ∠D = 90

∠PAD + ∠PDA = 90

∠APD = 90

∠SPQ = 90

Similarly, ∠PQR = 90, ∠QRS = 90

And ∠PSR = 90

Thus, PQRS is a quadrilateral each of whose angles is 90.

Hence, PQRS is a rectangle.

Given, ABCD is a parallelogram whose diagonals bisect each other at O.

Now, in ΔODP and ΔOBQ,

∠BOQ = ∠POD

∠OBQ = ∠ODP (AD||BC and BD is transversal)

OB = OD

OP = OQ

Hence, O bisect PQ.

Given, in a rectangle ABCD, diagonal BDl bisects ∠B.

Now, join AC.

In ΔBAD and ΔBCD,

∠ABD = ∠CBD

∠A = ∠C = 90

And BD = BD (common)

BADBCD

AB = BC

And AD = CD

But in rectangle opposite sides are equal,

AB = CD

And BC = AD

AB = BC = CD = DA

So, ABCD is a square.

Hence, Proved.

ABC is a triangle and D, E and F are mid-points of sides AB, BC and CA, respectively. Then,

AD = BD = AB

BE = EC = BC

And AF = CF = AC

Now, by mid-point theorem,

EF||AB and EF = AB = AD = BD

ED||AC and ED = AC = AF = CF

DF||BC and DF = BC = BE = CE

Now, in ΔADF and ΔEFD,

AD = EF

AF = DE

And DF = FD (common)

ADF EFD

Similarly, ΔDEF DEB

And ΔDEF CEF

Thus, ΔABC is divided into four congruent triangles.

Let ABCD be a trapezium in which AB||DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

Now, join CN and produce it to AB at E.

In ΔCDN and ΔEBN, we have,

DN = BN

∠DCN = ∠BEN

∠CDN = ∠EBN

ΔCDN EBN

DC = EB and CN = NE

Thus, in ΔCAE, the points M and N are the mid-points of AC and CE, respectively.

MN||AE

MN||AB||CD

Hence, proved.

Given, ABCD is a parallelogram.

BC = AD and BC||AD

Also, DC = AB and DC||AB

Since, P is mid-point of DC.

DP = PC = DC

Now, QC||AP and PC||AQ

APCQ is a parallelogram.

AQ = PC = DC = AB = BQ

Now, in ΔAQR and ΔBQC,

BQ = AQ

∠BQC = ∠AQR

And ∠BCQ = ∠ARQ

ΔAQR BQC

AR = BC

But BC = DA

AR = DA

Also, CQ = QR

Hence, Proved.