Congruent Triangles

Given,

The sides BA and CA have been produced, such that:

BA = AD

And, CA = AE

We have to prove that,

DE ‖ BC

Consider and , we have

BA = AD and CA = AE (Given)

∠BAC = ∠DAE (Vertically opposite angle)

So, by SAS congruence rule we have:

Therefore, BC = DE and

∠DEA = ∠BCA,

∠EDA = ∠CBA (By c.p.c.t)

Now, DE and BC are two lines intersected by a transversal DB such that,

∠DEA = ∠BCA,

i.e., Alternate angles are equal

Therefore, DE BC

Given that in Δ PQR,

PQ = QR

And, L, M, N are the mid points of the sides PQ, QR and RP respectively.

We have to prove that,

LN = MN

Here, we can observe that PQR is an isosceles triangle

PQ = QR

And, ∠QPR = ∠QRP (i)

And also, L and M are the mid points of PQ and QR respectively

PL = LQ =

QM = MR =

And, PQ = QR

PL = LQ = QM = MR = = (ii)

Now, in

LP = MR (From ii)

∠LPN = ∠MRN (From i)

PN = NR (N is the mid-point of PR)

Hence, By SAS theorem

, LN = MN (By c.p.c.t)

Given,

PQRS is a square and SRT is a equilateral triangle

To prove: (i) PT = QT

(ii) ∠TQR =15°

Proof: PQ = QR = RS = SP (As PQRS is a square, all sides will be equal) (i)

And, ∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90^o

And also,

SRT is an equilateral triangle

SR = RT = TS (ii)

And, ∠TSR = ∠SRT = ∠RTS = 60^o

From (i) and (ii)

PQ = QR = SP = SR= RT = TS (iii)

∠TSP = ∠TSR + ∠RSP

= 60^o + 90^o = 150^o

∠TRQ = ∠TRS + ∠SRQ

= 60^o + 90^o = 150^o

Therefore, ∠TSR = ∠TRQ = 150^o (iv)

Now, in and , we have

TS = TR (From iii)

∠TSP = ∠TRQ (From iv)

SP = RQ (From iii)

Therefore, By SAS theorem,

PT = QT (BY c.p.c.t)

In

QR = TR (From iii)

Hence, is an isosceles triangle.

Therefore, ∠QTR = ∠TQR (Angles opposite to equal sides)

Now,

Sum of angles in a triangle is 180^o

∠QTR + ∠TQR + ∠TRQ = 180^O

2∠TQR + 150^O = 180^O (From iv)

2∠TQR = 30^O

∠TQR = 15^O

Hence, proved

To prove: The medians of an equilateral triangle are equal.

Median = The line joining the vertex and mid-points of opposite sides.

Proof: Let Δ ABC be an equilateral triangle

AD, EF and CF are its medians.

Let,

AB = AC = BC = x

In BFC and CEB, we have

AB = AC (Sides of equilateral triangle)

AB = AC

BF = CE

∠ABC =∠ACB (Angles of equilateral triangle)

BC = BC (Common)

Hence, by SAS theorem, we have

Δ BFC ≅ Δ CEB

BE = CF (By c.p.c.t)

Similarly, AB = BE

Therefore, AD = BE = CF

Hence, proved

Given,

∠A =120°

AB = AC

We have to find ∠B and ∠C:

We can observe that Δ ABC is an isosceles triangle since AB = AC

∠B = ∠C (Angle opposite to equal sides are equal) [i]

We know that,

Sum of angles in a triangle is equal to 180^o

∠A + ∠B + ∠C = 180^o

∠A + ∠B + ∠B = 180^o

∠A + 2∠B = 180^o

120^o + 2∠B = 180^o

2∠B = 180^o – 120^o

2∠B = 60^o

∠B = 30^o

∠B = ∠C = 30^o

Consider Δ ABC,

We have,

∠B = 70°

And, AB = AC

Therefore, Δ ABC is an isosceles triangle.

∠B = ∠C (Angle opposite to equal sides are equal)

∠B = ∠C = 70^o

And ∠A + ∠B + ∠C = 180^o (Angles of triangle)

∠A + 70^o + 70^o = 180^o

∠A = 40^o

Consider an isosceles triangle ABC,

Such that:

AB = AC

Given,

Vertical ∠A is 100^o

To find: Base angle

Since, is isosceles,

∠B = ∠C (Angle equal to opposite sides)

And,

∠A + ∠B + ∠C = 180^o (Angles of triangle)

100^o + ∠B + ∠B = 180^o

∠B = 40^o

∠B = ∠C = 40^o

Given,

AB = AC

∠ACD = 105^o

Since, ∠BCD = 180^o (Straight angle)

∠BCA + ∠ACD = 180^o

∠BCA + 105^o = 180⁰

∠BCA = 75^o (i)

Now,

is an isosceles triangle

∠ABC = ∠ACB (Angle opposite to equal sides)

From (i), we have

∠ACB = 75^o

∠ABC = ∠ACB = 75^o

Sum of interior angle of triangle = 180^o

∠A +∠B +∠C =180°

∠A = 180° - 75° -75°

=30°

Therefore, ∠BAC = 30^o

To Find: Measure of each exterior angle of an equilateral triangle

Consider an equilateral triangle ABC.

We know that, for an equilateral triangle

AB = AC =CA

And, ∠ABC = ∠BCA = ∠CAB =

= 60^o (i)

Now, Extend side BC to D,

CA to E and AB to F

Here,

BCD is a straight line segment

∠BCD = Straight line segment = 180^o

∠BCA + ∠ACD = 180^o

60^o + ∠ACD = 180^o (From i)

∠ACD = 120^o

Similarly, we can find ∠EAB and ∠FBC also as 120^o because ABC is an equilateral triangle.

Therefore, ∠ACD = ∠EAB = ∠FBC = 120^o

Hence, the measure of each exterior angle of an equilateral triangle is 120^o

To prove: the exterior angles formed are equal to each other

i.e., ∠ADB =∠ACE

Proof: Let ABC be an isosceles triangle

Where BC is the base of the triangle and AB and AC are its equal sides.

∠ABC =∠ACB

∠B =∠C (Angle opposite to equal sides)

Now,

∠ADB +∠ABC = 180°

∠ACB +∠ACE =180°

∠ADB =180° -∠B

And

∠ACE=180°-∠C

∠ADB = 180° - ∠B

And

∠ACE=180°-∠B

∠ADB =∠ACE

Hence, proved

Consider the figure,

Given,

AB = AC

DB = DC

To find: Ratio ∠ABD =∠ACD

Now, are isosceles triangles

Since, AB = AC

And,

DB = DC

Therefore, ∠ABC = ∠ACB and,

∠DBC = ∠DCB (Angle opposite equal sides)

Now, consider ∠ABD: ∠ACD

(∠ABC - ∠DBC): (∠ACB - ∠DCB)

(∠ABC - ∠DBC): (∠ABC - ∠DBC) [Since, ∠ABC = ∠ACB and ∠DBC = ∠DCB]

1: 1

Therefore, ∠ABD: ∠ACD = 1:1

Given,

ABC is a right-angled triangle

∠A = 90°

And,

AB = AC

To find: ∠B and ∠C

Since, AB = AC

Therefore, ∠B = ∠C

And, Sum of angles in a triangle = 180^o

∠A + ∠B + ∠ C = 180^o

90^o + 2∠B = 180^o

2∠B = 90^o

∠B = 45^o

Hence, the measure of each angle of the equal angles of a right angle isosceles triangle is 45^o.

Consider the figure,

We have AB is a line segment

P, Q are the points on opposite sides on AB

Such that,

AP = BP (i)

AQ = BQ (ii)

To prove: PQ is perpendicular bisector of AB

Proof: Now, consider ,

AP = BP (From i)

AQ = BQ (From ii)

PQ = PQ (Common)

Therefore, By SSS theorem

(iii)

are isosceles triangles [From (i) and (ii)]

∠PAB = ∠PBA

And,

∠QAB = ∠QBA

Consider,

C is the point of intersection of AB and PQ

PA = PB [From (i)]

∠APC = ∠BPC [From (iii)]

PC = PC (Common)

By SAS theorem,

AC = CB

And, ∠PCA = ∠PCB (By c.p.c.t) (iv)

And also,

ACB is line segment

∠ACP + ∠BCP = 180^o

But, ∠ACP = ∠PCB

∠ACP = ∠PCB = 90^o (v)

We have,

AC = CB

C is the mid-point of AB

From (iv) and (v), we conclude that

PC is the perpendicular bisector of AB

Since, C is the point on line PQ, we can say that PQ is the perpendicular bisector of AB.

In the figure, given that:

RT = TS (i)

∠1 = 2∠2 (ii)

And,

∠4 = 2∠3 (iii)

To prove:

Let the point of intersection of RB and SA be denoted by O.

Since, RB and SA intersect at O.

∠AOR = ∠BOS (Vertically opposite angle)

∠1 = ∠4

2∠2 = 2∠3 [From (ii) and (iii)]

∠2 = ∠3 (iv)

Now, we have in

RT = TS

is an isosceles triangle

Therefore, ∠TRS = ∠TSI (v)

But, we have

∠TRS = ∠TRB + ∠2 (vi)

∠TSR = ∠TSA + ∠3 (vii)

Putting (vi) and (vii) in (v), we get

∠TRB + ∠2 = ∠TSA + ∠3

∠TRB = ∠TSA [From (iv)]

Now, in

RT = ST [From (i)

∠TRB = ∠TSA [From (iv)

∠RTB = ∠STA (Common angle)

By ASA theorem,

Given that,

Lines AB and CD intersect at O such that:

BC ‖ AD

And, BC = AD (i)

To prove: AB and CD bisect at O

Proof: In Δ AOD and Δ BOC

AD = BC [From (i)]

∠OBC =∠OAD (AD||BC and AB is transversal)

∠OCB =∠ODA (AD||BC and CD is transversal)

Therefore, by ASA theorem:

Δ AOD ≅ Δ BOC

OA = OB (By c.p.c.t)

And,

OD = OC (By c.p.c.t)

Hence, AB and CD bisect each other at O.

Given,

In isosceles Δ ABC,

BD and CE are bisectors of ∠B and ∠C

And,

AB = AC

To prove: BD = CE

Proof: In Δ BEC and Δ CDB, we have

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

BC = BC (Common)

∠BCE = ∠CBD (Since, ∠C = ∠B ∠C = ∠B ∠BCE = ∠CBD)

By ASA theorem, we have

Δ BEC ≅ Δ CDB

EC = BD (By c.p.c.t)

Hence, proved

Given that in two right angle triangles one side and acute angle of one are equal to the corresponding side and angle of the other.

We have to prove that the triangles are congruent.

Let us consider two right angle triangles. Such that,

∠B = ∠E = 90^o (i)

AB = DE (ii)

∠C = ∠F (iii)

Now, observe the two triangles ABC and DEF

∠C = ∠F (iv)

∠B = ∠E [From (i)]

AB = DE [From (ii)]

So, by AAS theorem, we have

Hence, proved

Given that the bisector of exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles.

Let, ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD ‖ BC

Let, ∠EAD = 1

∠DAC = 2

∠ABC = 3

∠ACB = 4

We have,

1 = 2 (Therefore, AD is the bisector of ∠EAC)

1 = 3 (Corresponding angles)

And,

2 = 4 (Alternate angles)

3 = 4 = AB = AC

Since, in two sides AB and AC are equal we can say that is isosceles.

Let be isosceles

Such that,

AB = BC

∠B = ∠C

Given, that vertex angle A is twice the sum of the base angles B and C.

i.e., ∠A = 2(∠B + ∠C)

∠A = 2(∠B + ∠B)

∠A = 2(2∠B)

∠A = 4∠B

Now,

We know that the sum of all angles of triangle = 180^o

∠A + ∠B + ∠C = 180^o

4∠B + ∠B + ∠B = 180^o (Therefore, ∠A = 4∠B, ∠C = ∠B)

6∠B = 180^o

∠B =

= 30^o

Since, ∠B = ∠C = 30^o

And, ∠A = 4∠B

= 4 * 30^o = 120^o

Therefore, the angles of the triangle are 120^o, 30^o, 30^o.

Given that PQR is a triangle

Such that,

PQ = PR

And, S is any point on side PQ and ST ‖ QR

We have to prove PS = PT

Since,

PQ = PR

PQR is isosceles

∠Q = ∠R

Or, ∠PQR = ∠PRQ

Now,

∠PST = ∠PQR

And,

∠PTS = ∠PRQ (Corresponding angles as ST ‖ QR)

Since,

∠PQR = ∠PRQ

∠PST = PTS

Now, in

∠PST = ∠PTS

Therefore, is an isosceles triangle

PS = PT

Given that in Δ ABC,

AB = AC and the bisectors of ∠B and ∠C intersect at O and M is a point on BO produced.

We have to prove ∠MOC = ∠ABC

Since,

AB = AC

Δ ABC is isosceles

∠B = ∠C

Or,

∠ABC = ∠ACB

Now,

BO and CO are bisectors of ∠ABC and ∠ACB respectively.

∠ABO = ∠OBC = ∠ACO = ∠OCB = ∠ABC = ∠ACB (i)

We have, in

∠OCB + ∠OBC + ∠BOC = 180^o (ii)

And also,

∠BOC + ∠COM = 180^o (iii) [Straight angle]

Equating (ii) and (iii), we get

∠OCB + ∠OBC + ∠BOC = ∠BOC + ∠COM

∠OBC + ∠OBC = ∠MOC

2∠OBC = ∠MOC

2(∠ABC) = ∠MOC [From (i)]

∠ABC = ∠MOC

Therefore, ∠MOC = ∠ABC

Given that P is the point on the bisector of an angle ∠ABC, and PQ ‖ AB

We have to prove that BPQ is isosceles

Since,

BP is the bisector of ∠ABC = ∠ABP = ∠PBC (i)

Now,

PQ ‖ AB

∠BPQ = ∠ABP (ii) [Alternate angles]

From (i) and (ii), we get

∠BPQ = ∠PBC

Or,

∠BPQ = ∠PBQ

Now, in

∠BPQ = ∠PBQ

is an isosceles triangle

Given to prove that each angle of the equilateral triangle is 60^o

Let us consider an equilateral triangle ABC

Such that,

AB = BC = CA

Now,

AB = BC

∠A = ∠C [i] (Opposite angles to equal sides are equal)

BC = AC

∠B = ∠A [ii[ (Opposite angles to equal sides are equal)

From [i] and [ii], we get

∠A = ∠B = ∠C [iii]

We know that,

Sum of all angles of triangles = 180^o

∠A + ∠B + ∠C = 180^o

∠A + ∠A + ∠A = 180^o

3∠A = 180^o

∠A =

= 60^o

Therefore, ∠A = ∠B = ∠C = 60^o

Hence, each angle of an equilateral triangle is 60^o.

Given that A, B, C of a triangle ABC are equal to each other.

We have to prove that, Δ ABC is equilateral.

We have,

∠A = ∠B = ∠C

Now,

∠A = ∠B

BC = AC (Opposite sides to equal angles are equal)

∠B = ∠C

AC = AB (Opposite sides to equal angles are equal)

From the above, we get

AB = BC = AC

Therefore,

Hence, proved

Given that, in ,

∠B = 2∠C and,

D is a mid-point on BC such that AD bisects ∠BAC and AB = CD.

We have to prove that, ∠BAC = 72^o

Now draw the angular bisector of ∠ABC, which meets AC in P

Join PD

Let, ∠ACB = y

∠B = ∠ABC = 2∠C = 2y

Also, ∠BAD = ∠DAC = x

∠BAC = 2x (Therefore, AD is the bisector of ∠BAC)

Now, in

∠CBP = y (Therefore, BP is the bisector of ∠ABC)

∠PCB = y

∠CBP = ∠PCB = y

Therefore, PC = BP

Consider , we have

∠ABP = ∠DCP = y

AB = DC (Given)

PC = BP (From above)

So, by SAS theorem, we have

Now,

∠BAP = ∠CDP

And, AP = DP (By c.p.c.t)

∠BAP = ∠CDP = 2x

Now, in

∠ABD + ∠BAD + ∠ADB = 180^o

∠ADB + ∠ADC = 180^o (Straight angle)

2x + 2y + y = 180^o (Therefore, ∠A = 2x, ∠B = 2y, ∠C = y)

2y + 3y = 180^o (Therefore, x = y)

5y = 180^o

y =

y = 36^o

Therefore, x = y = 36^o

Now,

∠A = ∠BAC = 2x = 2 * 36^o = 72^o

Therefore, ∠BAC = 72^o

Hence, proved

Given that ABC is a right angled triangle

Such that,

∠A = 90^o

And,

AB = AC

Since,

AB = AC

is also isosceles triangle

Therefore, we can say that is a right angled isosceles triangle.

∠C = ∠B

And,

∠A = 90^o

Now, we have

Sum of angles in a triangle = 180^o

∠A + ∠B + ∠C = 180^o

90^o + ∠B + ∠B = 180^o (From i)

90^o + 2∠B = 180^o

2∠B = 90^o

∠B = 45^o

Therefore, ∠B = ∠C = 45^o

Given, in the figure

AB = CD

And,

AD = BC

To prove: Δ ADC ≅ Δ CBA

Proof: Consider,

AB = CD (Given)

BC = AD (Given)

AC = AC (Common)

By SSS theorem,

Δ ADC ≅ Δ CBA

Hence, proved

Given that,

In Δ PQR

PQ = QR

And,

L, M, and N are the mid points of PQ, QR and RP respectively

To prove: LM = MN

Construction: Join L and M, M and N and N and L

Proof: We have,

PL = LQ, QM = MR and RN = NP

Since, L, M and N are mid points of PQ, QR and RP respectively.

And, also PQ = QR

PL = LQ = QM = MR = = (i)

Using mid-point theorem, we have

MN ‖ PQ

And,

MN = PQ = MN = PL = LQ (ii)

Similarly, we have

LN ‖ QR

And,

LN = QR = LN = QM = MR (iii)

From equations (i), (ii) and (ii), we have

PL = LQ = QM = MR = MN = LN

Therefore, LN = MN

Hence, proved

Given,

ABC is a triangle and D is the mid-point of BC

Perpendicular from D to AB and AC are equal.

To prove: Triangle is isosceles

Proof: Let DE and DF be perpendiculars from A on AB and AC respectively.

In order to prove that AB = AC, we will prove that Δ BDE ≅ Δ CDF.

In these two triangles, we have

∠BEF =∠CFD = 90°

BD = CD (Therefore, D is the mid-point of BC)

DE=DF (Given)

So, by RHS congruence criterion, we have

Δ BDE ≅ Δ CDF

∠B = ∠C (By c.p.c.t)

AC = AB (By c.p.c.t)

As opposite sides and opposite angles of the triangle are equal.

Therefore, Δ ABC is isosceles

Given that ABC is a triangle in which BE and CF are perpendiculars to the side AC and AB respectively.

Such that,

BE = CF

We have to prove that, is isosceles triangle.

Now, consider

We have,

∠BFC = ∠CEB = 90^o (Given)

BC = CB (Given)

CF = BE (Given)

So, by RHS congruence rule, we have

∠FBC = ∠ECB (By c.p.c.t)

∠ABC = ∠ACB (By c.p.c.t)

AC = AB (Opposite sides of equal angles are equal in a triangle)

Therefore, is isosceles.

Given that perpendiculars from any point within an angle on its arms are congruent.

We have to prove that it lies on the bisector of that angle.

Now, let us consider an ∠ABC and let BP be one of the arm within the angle.

Draw perpendicular PN and PM On the arms BC and BA

Such that,

They meet BC and BA in N and M respectively.

Now, in

We have,

∠BMP = ∠BNP = 90^o (Given)

BP = BP (Common)

MP = NP (Given)

So, by RHS congruence rule, we have

∠MBP = ∠NBP (By c.p.c.t)

BP is the angular bisector of ∠ABC

Hence, proved

Given that in figure,

AD ⊥ CD and CB ⊥ CD

And,

AQ = BP, DP = CQ

WE have to prove that,

∠DAQ = ∠CBP

Given that, DP = QC

Adding PQ on both sides, we get

DP + PQ = PQ + QC

DQ = PC (i)

Now consider , we have

∠ADQ = ∠BCP = 90^o (Given)

AQ = BP (Given)

And,

DQ = PC (From i)

So, by RHS congruence rule, we have

Now,

∠DAQ = ∠CBP (By c.p.c.t)

Hence, proved

Given that ABCD is a square, X and Y are points on the sides AD and BC respectively.

Such that,

AY = BX

We have to prove: BY = AX and ∠BAY = ∠ABX

Join B and X, A and Y

Since, ABCD is a square

∠DAB = ∠CBA = 90^o

∠XAB = ∠YBA = 90^o (i)

Now, consider

We have,

∠XAB = ∠YBA = 90^o [From (i)]

BX = AY (Given)

AB = BA (Common side)

So, by RHS congruence rule, we have

(By c.p.c.t)

(i) False: Sides opposite to equal angles of a triangle are equal.

(ii) True: Since, the sides are equal, the corresponding opposite angles must be equal.

(iii) True: Since, all the three angles of an equilateral triangle are equal and sum of the three angles is 180^o, so each angle will be equal to = 60^o

(iv) False: Here, the altitude from the vertex is also the perpendicular bisector of the opposite side. Here the triangle must be isosceles and may be an equilateral triangle.

(v) True: Since, it is an isosceles triangle, the length of bisector of the two angles are equal.

(vi) False: The angular bisector of the vertex angle is also a median. The triangle must be an isosceles and an equilateral triangle.

(vii) False: Since, two sides are equal the triangle must be an isosceles triangle. The two altitudes corresponding to two equal sides must be equal.

(viii) False: The two right triangles may or may not be congruent.

(ix) True: According to RHS congruence the given statement is true.

(i) Sides opposite to equal angles of a triangle are equal

(ii) Sides opposite to equal angles of a triangle are equal

(iii) In an equilateral triangle all angles are equal

(iv) In a Δ ABC, if ∠A=∠C, then AB = BC

(v) If altitudes CE and BF of a triangle ABC are equal, then, AB = AC

(vi) In an isosceles triangle ABC with AB=AC, if BD and CE are its altitudes, then BD is equal to CE

(vii) In right triangles ABC and DEF, if hypotenuse AB=EF and side AC=DE, then Δ ABC ≅ ΔEFD

Given that in Δ ABC

∠A = 40^o and ∠B = 60^o

We have to find shortest and longest side.

We know that,

Sum of angles of triangle = 180^o

∠A + ∠B + ∠C = 180^o

40^o + 60^o + ∠C = 180^o

100^o + ∠C = 180^o

∠C = 180^o – 100^o

= 80^o

Now,

40^o < 60^o < 80^o

∠A < ∠B < ∠C

∠C is greater angle and ∠A is smaller angle.

As, ∠A < ∠B < ∠C

BC < AC < AB (Therefore, side opposite to greater angle is larger and side opposite to smaller angle is smaller)

Therefore, AB is longest and BC is smallest or shortest side.

Given that in ,

∠B = ∠C = 45^o

We have to find longest side.

We know that,

Sum of angles in a triangle = 180^o

∠A + ∠B + ∠C = 180^o

∠A + 45^o + 45^o = 180^o

∠A + 90^o = 180^o

∠A = 90^o

Therefore, BC is the longest side because side opposite to greater angle is larger.

Given that in , side AB is produced to D so that BD = BC and ∠B = 60^o, ∠A = 70^o

We have to prove that,

(i) AD > CD

And, (ii) AD > AC

First join C and D

Now,

In ,

∠A + ∠B + ∠C = 180^o (Sum of all angles of triangle)

∠C = 180^o – 70^o – 60^o

= 50^o

∠C = 50^o

∠ACB = 50^o (i)

And also in ,

∠DBC = 180^o - ∠ABC (Therefore, ∠ABD is straight angle)

= 180^o – 60^o

= 120^o

BD = BC (Given)

∠BCD = ∠BDC (Therefore, angle opposite to equal sides are equal)

Now,

∠DBC + ∠BCD + ∠BDC = 180^o (Sum of all sides of triangle)

120^o + ∠BCD + ∠BCD = 180^o

2∠BCD = 180^o – 120^o

2∠BCD = 60^o

∠BCD = 30^o

Therefore, ∠BCD = ∠BDC = 30^o (ii)

Now, consider ,

∠BAC = ∠DAC = 70^o (Given)

∠BDC = ∠ADC = 30^o [From (ii)]

∠ACD = ∠ACB + ∠BCD

= 50^o + 30^o [From (i) and (ii)]

= 80^o

Now,

∠ADC < ∠DAC < ∠ACD

AC < DC < AD (Therefore, side opposite to greater angle is longer and smaller angle is smaller)

AD > CD

And,

AD > AC

Hence, proved

We have,

∠ACD > ∠DAC

And,

∠ACD > ∠ADC

AD > DC

And,

AD > AC (Therefore, side opposite to greater angle is longer and smaller angle is smaller)

Given, Length of sides are 2cm, 3cm and 7cm.

We have to check whether it is possible to draw a triangle with the given length of sides.

We know that,

A triangle can be drawn only when the sum of any two sides is greater than the third side.

Here,

2 + 3>7

So, the triangle does not exist.

Given: ∠B = 35°

∠C = 65°

The bisector of ∠ABC meets BC in P

We have to arrange AP, BO and CP in descending order

In Δ ACP, we have

∠ACP > ∠CAP

AP > CP (i)

In Δ ABP, we have

∠BAP > ∠ABP

BP > AP (ii)

From (i) and (ii), we have

BP > AP > CP

Given that, O is any point in the interior of

We have to prove:

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > (AB + BC + CA)

We know that,

In a triangle sum of any two sides is greater than the third side.

So, we have

In

AB + BC > AC

BC + AC > AB

AC + AB > BC

In ,

OB + OC > BC (i)

In ,

OA + OC > AC (ii)

In ,

OA + OB > AB (iii)

Now, extend BO to meet AC in D.

In , we have

AB + AD > BD

AB + AD > BO + OD (iv) [Therefore, BD = BO + OD]

Similarly,

In , we have

OD + DC > OC (v)

(i) Adding (iv) and (v), we get

AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC (vi)

Similarly, we have

BC + BC > OA + OC (vii)

And,

CA + CB > OA + OB (viii)

(ii) Adding (vi), (vii) and (viii), we get

AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB

2AB + 2BC + 2CA > 2OA + 2OB + 2OC

2 (AB + BC + CA) > 2 (OA + OB + OC)

AB + BC + CA > OA + OB + OC

(iii) Adding (i), (ii) and (iii), we get

OB + OC + OA + OC + OA + OB > BC + AC + AB

2OA + 2OB + 2OC > AB + BC + CA

2 (OA + OB + OC) > AB + BC + CA

Therefore, (OA + OB + OC) > (AB+ BC + CA)

Given: A in which AD perpendicular BC and BE perpendicular AC and CF perpendicular AB.

To prove: AD + BE + CF < AB + BC + AC

Proof: We know that all the segments that can be drawn into a given line, from a point not lying on it, perpendicular distance i.e. the perpendicular line segment is the shortest. Therefore,

AD perpendicular BC

AB > AD and AC > AD

AB + AC > 2AD (i)

Similarly,

BE perpendicular AC

BA > BE and BC > BE

BA + BC > 2BE (ii)

And also

CF perpendicular AB

CA > CF and CB > CF

CA + CB > 2CF (iii)

Adding (i), (ii) and (iii), we get

AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF

2AB + 2BC + 2CA > 2 (AD + BE + CF)

2 (AB + BC + CA) > 2 (AD + BE + CF)

AB + BC + CA > AD + BE + CF

The perimeter of the triangle is greater than the sum of its altitudes.

Hence, proved

Given: Let ABCD is a quadrilateral with AC and BD as its diagonals

To Prove: Sum of all the sides of a quadrilateral is greater than the sum of its diagonals

Proof: Consider a quadrilateral ABCD where AC and BD are the diagonals

AB+BC > AC (i) (Sum of two sides is greater than the third side)

AD+DC > AC (ii)

AB+AD > BD (iii)

DC+BC > BD (iv)

Adding (i), (ii), (iii), and (iv)

AB+BC+AD+DC+AB+AD+DC+BC > AC+AC+BD+BD

2(AB+BC+CD+DA) > 2(AC+BD)

AB+BC+CD+DA > AC+BC

Hence, proved that the Sum of all the sides of a quadrilateral is greater than the sum of its diagonals

Given to prove,

(i) CD + DA + AB + BC > 2AC

(ii) CD+DA +AB > BC

From the given figure,

We know that,

In a triangle sum of any two sides is greater than the third side.

(i) So,

In , we have

AB + BC > AC (1)

In , we have

CD + DA > AC (2)

Adding (1) and (2), we get

AB + BC + CD + DA > AC + AC

CD + DA + AB + BC > 2 AC

(ii) Now, in , we have

CD + DA > AC

Add AB on both sides, we get

CD + DA +AB > AC + AB > BC

CD + DA + AB > BC

Hence, proved

(i) False: Sum of three sides of a triangle is greater than sum of its three altitudes.

(ii) True

(iii) True

(iv) False: The difference of any two sides of a triangle is less than the third side.

(v) True: The side opposite to greater angle is longer and smaller angle is shorter in a triangle.

(vi) True: The perpendicular distance is the shortest distance from a point to a line not containing it.

(i) In a right triangle the hypotenuse is the largest side.

(ii) The sum of three altitudes of a triangle is less than its perimeter.

(iii) The sum of any two sides of a triangle is greater than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.

(v) Difference of any two sides of a triangle is less than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

By c.p.c.t. that is corresponding part of congruent triangles, the pair of equal angles are:

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Since, by corresponding part of congruent triangle AC of ABC is equal to the LM of LKM.

No, the two triangles are not necessarily congruent in the given case as knowing only angle-angle-angle (AAA) does not work because it can produce similar but not congruent triangles.

AB and AC are the equal sides of equilateral triangle ABC because only then after reverting it to form ACB, the two triangles can be proved congruent.

Since, BC and PQ are the non c.p.c.t. part of the two given triangles, hence we cannot judge them to be equal or not from the given information about them.

Yes, the given triangles are congruent as AC=DE, BC = DF and ∠D is equal to ∠C. Hence, By SAS theorem triangle ABC is congruent to triangle EDF.

Since, two adjacent sides and the so formed angle with the two sides are shown equal while proving them congruent. Hence, by S.A.S. theorem the two triangles can be proved congruent.

Yes, the given triangles are congruent as AB= AD, BC= CD and AC is a common side. Hence, by SSS theorem triangle ABC is congruent to triangle ADC.

Since, the two adjacent angles and the contained side are shown equal while proving the two triangles congruent. Hence, by A.S.A. theorem the two triangles can be proved congruent.

Yes, the two given triangles are congruent because AC = CE, BC= CD

And ∠B = ∠D= 90°.

Therefore by SSA criteria triangle ABC is congruent to triangle CDE.

Since, by corresponding part of congruent triangle ED of EFD is equal to the PR of PQR.

Given,

ABC is an isosceles triangle

AB = AC

BE and CF are two medians

To prove: BE = CF

Proof: In

CE = BF (Since, AC = AB = = AB = CE = BF)

∠ECB = ∠FBC (Angle opposite to equal sides are equal)

BC = BC (Common)

Therefore, By SAS theorem

BEC

(By c.p.c.t)

Since, by corresponding part of congruent triangle ∠E of EFD is equal to the ∠P of PQR.

Let ABC is an equilateral triangle

We know that all the angles in an equilateral triangle are equal.

Therefore,

∠A = ∠B = ∠C (i)

Now,

∠A + ∠B + ∠C = 180^o (Sum of all angles of a triangle)

∠A + ∠A + ∠A = 180^o

3∠A = 180^o

∠A =

= 60^o

Hence, the measure of each angle of an equilateral triangle is 60^o.

∠ACB + ∠ACD = 180°

On solving we get,

∠ACB = 80°

∠ABC = ∠ACB = 80° (Angles opposite to equal sides are equal)

In ABC,

∠A + ∠B + ∠C= 180°

∠A + 80° + 80° = 180°

∠A = 20°

Given: An equilateral triangle CDE is on side CD of square ABCD

To prove:

Proof: ∠EDC = ∠DCE = ∠CED = 60^o (Angles of equilateral triangle)

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90^o (Angles of square)

∠EDA = ∠EDC + ∠CDA

= 60^o + 90^o

= 150^o (i)

Similarly,

∠ECB = 150^o (ii)

In

ED = EC (Sides of equilateral triangle)

AD = BC (Sides of square)

∠EDA = ∠ECB [From (i) and (ii)]

Therefore, By SAS theorem

Let the base angles be x each,

Vertex angle = 2(x + x) = 4x

Now, since the sum of all the angles of a triangle​ is 180°

x + x + 4x = 180°

6x = 180°

x = 30°

Therefore, vertex angle= 4x = 120°

In APB,

AP is a median

Angle APB is greater than angle ABP

So angle opposite to greater side is longer

Therefore, AB is greater than AP

Hence proved that in triangle APB the side of the triangle is more than the median or altitude AP of the triangle.

Similarly, in the same manner the other triangles can also be proved.

Given,

∠B = ∠C

AB = AC

To Prove: BQ = CP

Proof: In ABQ and ACP

∠B = ∠C (Given)

AB = AC (Given)

∠A = ∠A (Common)

Hence, by A.S.A. Theorem

ABQ ACP

BQ = CP (By c.p.c.t)

Since, the so formed triangle divides the complete triangle ABC into four congruent triangles.

Since the two triangles with two adjacent sides and an angle adjacent to any one side among them are shown equal, then the two triangles will be similar but not necessarily congruent.

In ABC and A'B'C',

AB= A'B' (Given)

∠B = ∠B'(Given)

BC = B'C' (Given)

Hence, by S.A.S. theorem,

ABC A'B'C'

Therefore,

By c.p.c.t

∠A = ∠A'

3x = 2x +20

x = 20°

Therefore angle A' = 2x +20

= 60°

In ABC,

∠A + ∠B + ∠C= 180

80° + 40°+ ∠C = 180°

∠C = 60°

Now, by c.p.c.t

AB = DF = 5cm

And

∠C = ∠E = 60°

In ABD and FEC,

AB= FE (Given)

∠B = ∠E (Each 90°)

BC = DE (Given)

Add CD both sides, we get

BD = EC

Therefore, by S.A.S. theorem,

ABD FEC

∠EAP = ∠BCA (Corresponding angles)

∠BCA = 70°

∠CBA = ∠BCA (Angles opposite to equal sides are equal)

∠CBA = 70°

Now,

∠ABD + ∠CBA = 180°

∠ABD + 70 = 180°

∠ABD = 110°

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

In ABC,

∠A + ∠B+ ∠C= 180°

∠A = 76°

Now,

∠BAC= ∠ACD (Alternate angles)

∠ACD = 76°

Now,

∠ACD + ∠ECD= 180°

x = 180- 76°

x = 104°

∠Z = 90° (Angle of square)

Therefore, AZD is a right angle triangle,

By Pythagoras theorem,

AD² = AZ² + ZD²

AD² = 2² + (2+3)²

AD² = 4 + 25

AD = √29

In AXB, with​X as right angle,

By Pythagoras theorem,

AB² = AX² + XB²

XB² = 29-4

XB = 5

BY = XB + XY

= 5 + 2

= 7cm

For an isosceles triangle ∠ABC = ∠ACB = 35°

Let the ∠ADB be x

Then,

∠ADC = 180^o - x

As AD is median so BD = CD

And for isosceles triangle AB = AC

So,

= = 1

By angle bisector theorem,

∠BAD = ∠CAD = y (Let)

For BAD

35 + x + y =180 (i)

For DAC

35 + 180 – x + y = 180 (ii)

35 + y = x

Therefore,

35 + 34 + y + y = 180^o

2y + 70 = 180^o

2y = 100^o

y = 55^o

Therefore,

∠BAD = ∠CAD = 55^o

Given that ABC

BE is bisector of ∠Band AD is bisector of ∠BAC

∠B = 2 ∠C

By exterior angle theorem in triangle ADC

∠ADB = ∠DAC + ∠C (i)

In ADB,

∠ABD + ∠BAD + ∠ADB = 180^o

2 ∠C + ∠BAD + ∠DAC + ∠C = 180^o [From (i)]

3 ∠C + ∠BAC = 180^o

∠BAC = 180^o – 3 ∠C (ii)

Therefore,

AB = CD

∠C = ∠DAC

∠C = 1/2 ∠BAC (iii)

Putting value of Angle C in (ii), we get

∠BAC = 180^o – 1/2 ∠BAC

∠BAC + ∠BAC = 180^o

∠BAC = 180^o

∠BAC =

= 72^o

∠BAC = 72^o

In using Pythagoras theorem, we get

AB² + BC² = AC²

9 + BC² = 25

BC = 4 cm

In

∠BAC = ∠CAD (Therefore, AC is bisector of ∠A)

∠B = ∠D = 90^o

∠ABC + ∠BCA + ∠CAB = 180^o

∠CAD + ∠ADC + ∠DCA = 180^o

∠ABC + ∠BCA + ∠CAB = ∠CAD + ∠ADC + ∠DCA

∠BCA = ∠DCA (i)

In

∠CAB = ∠CAD (Therefore, AC is bisector of ∠A)

∠BCA = ∠DCA [From (i)

AC = AC (Common)

By ASA theorem, we have

BC = CD (By c.p.c.t)

CD = 4cm