Isosceles Triangles [Including Inequalities] — Exercise 10(A)

In the given figure, ∠B = ∠C and ∠BAD = ∠CAD, then :

1. AB = AC

2. AB ≠ AC

3. ∠ADB ≠ ∠ADC

4. ∠ADB ≠ 90°

Answer

We know that,

Sides opposite to equal angles are equal.

Since, ∠B = ∠C

∴ AC = AB.

Hence, Option 1 is the correct option.

In the given figure, AD ⊥ BC and AB = AC, then :

1. △ ABD ≇ △ ACD

2. BD = CD

3. ∠BAC = 90°

4. ∠CAD = 45°

Answer

We know that,

Angles opposite to equal sides are equal.

Since, AB = AC,

∴ ∠C = ∠B.

In △ ABD and △ ACD,

⇒ AB = AC (Given)

⇒ AD = AD (Common side)

⇒ ∠B = ∠C (Proved above)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangles are equal.

∴ BD = CD.

Hence, Option 2 is the correct option.

In the given figure, AD = BD, then angle ACD is :

1. 43°

2. 22°

3. 65°

4. 28°

Answer

Given,

AD = BD

In △ ABD,

⇒ ∠BAD = ∠ABD = 65° (Angles opposite to equal sides are equal)

In △ ABC,

⇒ ∠BAC = ∠BAD + ∠DAC = 65° + 22° = 87°.

By angle sum property of triangle,

⇒ ∠BAC + ∠ACB + ∠CBA = 180°

⇒ 87° + ∠ACB + 65° = 180°

⇒ ∠ACB + 152° = 180°

⇒ ∠ACB = 180° - 152° = 28°.

From figure,

⇒ ∠ACD = ∠ACB = 28°.

Hence, Option 4 is the correct option.

In the given figure; BE = DC, then :

1. AD = DC

2. AE = BE

3. AD = AE

4. ∠ABE = ∠DAC

Answer

From figure,

⇒ AB = AC

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

Given,

⇒ BE = DC

⇒ BD + DE = DE + EC

⇒ BD = EC.

In △ ABD and △ AEC,

⇒ BD = EC (Proved above)

⇒ AB = AC (Given)

⇒ ∠B = ∠C (Proved above)

∴ △ ABD ≅ △ AEC (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ AD = AE.

Hence, Option 3 is the correct option.

In △ ABC and △ PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q; then triangles are :

1. isosceles but not congruent

2. isosceles and congruent

3. congruent but not isosceles

4. neither isosceles nor congruent.

Answer

△ ABC and △ PQR are shown below:

In △ ABC,

⇒ AB = AC (Given)

∴ ∠C = ∠B = x (let) (Angles opposite to equal sides are equal).

Given,

⇒ ∠P = ∠C

∴ ∠P = x.

⇒ ∠Q = ∠B

∴ ∠Q = x.

In △ PQR,

⇒ ∠P = ∠Q = x

⇒ QR = PR (Sides opposite to equal angles are equal).

∴ ABC and PQR are isosceles triangles.

Hence, Option 1 is the correct option.

In the figure alongside,

AB = AC

∠A = 48° and

∠ACD = 18°.

Show that : BC = CD.

Answer

In △ ABC,

⇒ AB = AC (Given)

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

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 48° + x + x = 180°

⇒ 48° + 2x = 180°

⇒ 2x = 180° - 48°

⇒ 2x = 132°

⇒ x = $\dfrac{132°}{2}$ = 66°.

∴ ∠B = ∠C = 66°.

From figure,

⇒ ∠DCB = ∠C - ∠ACD = 66° - 18° = 48°.

In △ BDC,

By angle sum property of triangle,

⇒ ∠BDC + ∠DCB + ∠CBD = 180°

⇒ ∠BDC + 48° + 66° = 180°

⇒ ∠BDC + 114° = 180°

⇒ ∠BDC = 180° - 114° = 66°.

Since, ∠BDC = ∠CBD

∴ BC = CD (Sides opposite to equal angles are equal).

Hence, proved that BC = CD.

Calculate :

(i) ∠ADC

(ii) ∠ABC

(iii) ∠BAC

Answer

(i) Since, DCE is a straight line.

∴ ∠ACD + ∠ACE = 180°

⇒ ∠ACD + 130° = 180°

⇒ ∠ACD = 180° - 130° = 50°.

In △ ADC,

⇒ AD = DC (Given)

⇒ ∠DAC = ∠ACD = 50° (Angles opposite to equal sides are equal.)

By angle sum property of triangle,

⇒ ∠ADC + ∠DAC + ∠ACD = 180°

⇒ ∠ADC + 50° + 50° = 180°

⇒ ∠ADC + 100° = 180°

⇒ ∠ADC = 180° - 100° = 80°.

Hence, ∠ADC = 80°.

(ii) Since, BDC is a straight line.

∴ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + 80° = 180°

⇒ ∠ADB = 180° - 80° = 100°.

In △ ABD,

⇒ AD = BD (Given)

⇒ ∠DBA = ∠BAD = x (let).

By angle sum property of triangle,

⇒ ∠BAD + ∠ADB + ∠DBA = 180°

⇒ x + 100° + x = 180°

⇒ 2x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

⇒ ∠DBA = ∠BAD = 40°.

From figure,

⇒ ∠ABC = ∠DBA = 40°.

Hence, ∠ABC = 40°.

(iii) From figure,

⇒ ∠BAC = ∠BAD + ∠DAC = 40° + 50° = 90°.

Hence, ∠BAC = 90°.

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate :

(i) ∠CDE

(ii) ∠DCE

Answer

Since, FAC is a straight line.

∴ ∠BAC + ∠FAB = 180°

⇒ ∠BAC + 128° = 180°

⇒ ∠BAC = 180° - 128° = 52°.

In △ ABC,

⇒ ∠A = 52°

⇒ AB = AC (Given)

∴ ∠B = ∠C = x (let) [Angles opposite to equal sides are equal].

By angle, sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 52° + x + x = 180°

⇒ 52° + 2x = 180°

⇒ 2x = 180° - 52°

⇒ 2x = 128°

⇒ x = $\dfrac{128°}{2}$ = 64°

⇒ ∠B = ∠C = 64°.

In △ DBC,

⇒ BC = CD (Given)

⇒ ∠BDC = ∠DBC = 64° (Angles opposite to equal sides are equal)

From figure,

⇒ ∠ADE = ∠ABC = 64° (Corresponding angles are equal)

Since, ADB is a straight line.

∴ ∠ADE + ∠CDE + ∠BDC = 180°

⇒ 64° + ∠CDE + 64° = 180°

⇒ ∠CDE + 128° = 180°

⇒ ∠CDE = 180° - 128° = 52°.

Hence, ∠CDE = 52°.

(ii) Since,

DE || BC

⇒ ∠DCB = ∠CDE = 52° (Alternate angles are equal)

From figure,

⇒ ∠DCE = ∠ECB - ∠DCB

= 64° - 52° = 12°.

Hence, ∠DCE = 12°.

Calculate x :

(i)

(ii)

Answer

(i) In △ ABC,

⇒ BC = AC (Given)

∴ ∠BAC = ∠CBA = 37° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ 37° + 37° + ∠ACB = 180°

⇒ ∠ACB + 74° = 180°

⇒ ∠ACB = 180° - 74° = 106°.

From figure,

Since, BCD is a straight line,

∴ ∠ACB + ∠ACD = 180°

⇒ 106° + ∠ACD = 180°

⇒ ∠ACD = 180° - 106° = 74°.

In △ ACD,

⇒ CD = AD (Given)

∴ ∠CAD = ∠ACD = 74° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠ACD + ∠ADC = 180°

⇒ 74° + 74° + x = 180°

⇒ 148° + x = 180°

⇒ x = 180° - 148° = 32°.

Hence, x = 32°.

(ii) In △ ABC,

⇒ BC = AC (Given)

∴ ∠BAC = ∠CBA = 50° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ 50° + 50° + ∠ACB = 180°

⇒ ∠ACB + 100° = 180°

⇒ ∠ACB = 180° - 100° = 80°.

From figure,

Since, BCD is a straight line,

∴ ∠ACB + ∠ACD = 180°

⇒ 80° + ∠ACD = 180°

⇒ ∠ACD = 180° - 80° = 100°.

In △ ACD,

⇒ CD = AC (Given)

∴ ∠CAD = ∠ADC = x (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠ACD + ∠ADC = 180°

⇒ x + x + 100° = 180°

⇒ 2x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

Hence, x = 40°.

In the figure, given below, AB = AC. Prove that : ∠BOC = ∠ACD.

Answer

In △ ABC,

⇒ AB = AC (Given)

∴ ∠C = ∠B = x (let)

From figure,

OB and OC bisects angle ∠B and ∠C.

∴ ∠OBC = $\dfrac{∠B}{2} = \dfrac{x}{2}$ and ∠OCB = $\dfrac{∠C}{2} = \dfrac{x}{2}$.

In △ BOC,

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ $\dfrac{x}{2} + \dfrac{x}{2}$ + ∠BOC = 180°

⇒ x + ∠BOC = 180°

⇒ ∠BOC = 180° - x.

From figure,

⇒ ∠OCA = $\dfrac{x}{2}$ (As OC is bisector of ∠C)

Since, BCD is a straight line.

∴ ∠OCB + ∠OCA + ∠ACD = 180°

⇒ $\dfrac{x}{2} + \dfrac{x}{2}$ + ∠ACD = 180°

⇒ x + ∠ACD = 180°

⇒ ∠ACD = 180° - x.

∴ ∠BOC = ∠ACD.

Hence, proved that ∠BOC = ∠ACD.

In the figure given below, LM = LN; angle PLN = 110°. Calculate :

(i) ∠LMN

(ii) ∠MLN

Answer

(i) In quadrilateral PQNL,

By angle sum property of quadrilateral,

⇒ ∠QPL + ∠PLN + ∠LNQ + ∠NQP = 360°

⇒ 90° + 110° + ∠LNQ + 90° = 360°

⇒ ∠LNQ + 290° = 360°

⇒ ∠LNQ = 360° - 290° = 70°.

From figure,

⇒ ∠LNM = ∠LNQ = 70°.

In triangle LMN,

⇒ LN = LM (Given)

⇒ ∠LMN = ∠LNM = 70°.

Hence, ∠LMN = 70°.

(ii) In triangle LMN,

By angle sum property of triangle,

⇒ ∠LMN + ∠LNM + ∠MLN = 180°

⇒ 70° + 70° + ∠MLN = 180°

⇒ ∠MLN + 140° = 180°

⇒ ∠MLN = 180° - 140° = 40°.

Hence, ∠MLN = 40°.

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°. Find :

(i) ∠DCB

(ii) ∠CBD

Answer

Isosceles triangle ABC is shown in the figure below:

(i) In △ ACD and △ BCD,

⇒ ∠CAD = ∠CBD (Since, AC = BC and angles opposite to equal sides are equal.)

⇒ AD = BD (CD bisects AB)

⇒ AC = BC (Given)

∴ △ ACD ≅ △ BCD (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠ADC = ∠BDC = x (let)

Since, ADB is a straight line.

∴ ∠ADC + ∠BDC = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

∴ ∠ADC = ∠BDC = 90°.

In △ BDC,

⇒ ∠BDC + ∠DCB + ∠CBD = 180°

⇒ 90° + ∠DCB + 55° = 180°

⇒ 145° + ∠DCB = 180°

⇒ ∠DCB = 180° - 145° = 35°.

Hence, ∠DCB = 35°.

(ii) From part (i), we get :

⇒ ∠CBD = 55°.

Hence, ∠CBD = 55°.

Find x :

Answer

In △ ADC,

⇒ AC = AD (Given)

⇒ ∠ACD = ∠ADC = 42° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ACD + ∠ADC + ∠CAD = 180°

⇒ 42° + 42° + ∠CAD = 180°

⇒ ∠CAD + 84° = 180°

⇒ ∠CAD = 180° - 84° = 96°.

Since, BCD is a straight line.

From figure,

⇒ ∠ACD + ∠ACB = 180°

⇒ 42° + ∠ACB = 180°

⇒ ∠ACB = 180° - 42° = 138°.

In △ ABC,

⇒ AC = BC (Given)

⇒ ∠ABC = ∠CAB = z (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ABC + ∠CAB + ∠ACB = 180°

⇒ z + z + 138° = 180°

⇒ 2z + 138° = 180°

⇒ 2z = 180° - 138°

⇒ 2z = 42°

⇒ z = $\dfrac{42°}{2}$ = 21°.

⇒ ∠ABC = ∠CAB = 21°.

Since, EAD is a straight line.

From figure,

⇒ ∠EAB + ∠CAB + ∠CAD = 180°

⇒ x + 21° + 96° = 180°

⇒ x + 117° = 180°

⇒ x = 180° - 117° = 63°.

Hence, x = 63°.

In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.

Answer

In △ ABD and △ CBD,

⇒ ∠ABD = ∠CBD (Since, BD bisects ∠B)

⇒ BD = BD (Common side)

⇒ ∠BDA = ∠BDC (Both equal to 90°)

∴ △ ABD ≅ △ CBD (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ AB = BC and AD = CD

Considering, AB = BC

∴ 3x + 1 = 5y - 2

⇒ 3x + 1 + 2 = 5y

⇒ 5y = 3x + 3

⇒ y = $\dfrac{3x + 3}{5}$ ..........(1)

Considering, AD = CD

∴ x + 1 = y + 2

⇒ y = x + 1 - 2

⇒ y = x - 1 ...........(2)

Equating equations (1) and (2), we get :

⇒ x - 1 = $\dfrac{3x + 3}{5}$

⇒ 5(x - 1) = 3x + 3

⇒ 5x - 5 = 3x + 3

⇒ 5x - 3x = 3 + 5

⇒ 2x = 8

⇒ x = $\dfrac{8}{2}$ = 4.

Substituting value of x in equation (2), we get :

⇒ y = 4 - 1 = 3.

Hence, x = 4 and y = 3.

In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.

Answer

Given,

⇒ ∠A : ∠B = 8 : 5

Let ∠A = 8x and ∠B = 5x.

In △ ABC,

⇒ AB = AC (Given)

⇒ ∠C = ∠B = 5x (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 8x + 5x + 5x = 180°

⇒ 18x = 180°

⇒ x = $\dfrac{180}{18}$ = 10°.

⇒ ∠A = 8x = 8 × 10° = 80°.

Hence, ∠A = 80°.

In triangle ABC; ∠A = 60°, ∠C = 40° and bisector of angle ABC meets side AC at point P. Show that BP = CP.

Answer

In △ ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 60° + ∠B + 40° = 180°

⇒ ∠B + 100° = 180°

⇒ ∠B = 180° - 100° = 80°.

Since, BP is bisector of ∠B,

∴ ∠PBC = $\dfrac{∠B}{2} = \dfrac{80°}{2}$ = 40°.

∵ ∠PBC = ∠PCB (Both equal to 40°)

∴ CP = BP (Sides opposite to equal angles are equal)

Hence, proved that BP = CP.