Triangles — Exercise 9.3

ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Answer

Given,

AB = AC.

We know that angles opposite to equal sides of an isosceles triangle are equal.

∴ ∠B = ∠C = x°.

Sum of angles of triangle = 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 90° + x° + x° = 180°

⇒ 90° + 2x° = 180°

⇒ 2x° = 90°

⇒ x° = 45°.

Hence, ∠B = ∠C = 45°.

Show that the angles of an equilateral triangle are 60° each.

Answer

In equilateral triangle all sides are equal.

AB = BC = CA

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

Sum of angles of triangle = 180°

∠A + ∠B + ∠C = 180°

x° + x° + x° = 180°

3x° = 180°

x = 60°.

Hence, proved that angles of an equilateral triangle are 60° each.

Show that every equiangular triangle is equilateral.

Answer

Let ABC be a triangle with, ∠A = ∠B = ∠C.

Considering, ∠A = ∠B

We get, BC = AC (Sides opposite to equal angles are equal.)

Considering, ∠B = ∠C

We get, AB = BC (Sides opposite to equal angles are equal.)

∴ AB = BC = AC.

Hence, proved that every equiangular triangle is equilateral.

In the following figure, find the value of x:

Answer

Since, △ABC is an isosceles triangle with AB = AC.

∴ ∠B = ∠C = y°

Sum of angles of triangle = 180°.

⇒ 50° + y° + y° = 180°

⇒ 50° + 2y° = 180°

⇒ 2y° = 180° - 50° = 130°.

⇒ y° = 65°.

From figure,

⇒ x° + y° = 180°

⇒ x° + 65° = 180°

⇒ x° = 180° - 65° = 115°.

Hence, x = 115°.

In the following figure, find the value of x:

Answer

From figure,

△PRS is an isosceles triangle with PR = SR.

∴ ∠S = ∠RPS = 30°.

∠QPS = ∠QPR + ∠RPS = 52° + 30° = 82°.

Since, sum of angles of triangle = 180°.

Considering △PQS we get,

⇒ ∠P + ∠Q + ∠S = 180°

⇒ 82° + x° + 30° = 180°

⇒ x° + 112° = 180°

⇒ x° = 180° - 112° = 68°.

Hence, x = 68°.

In the following figure, find the value of x:

Answer

△BDC is an isosceles triangle with BD = DC,

∴ ∠DBC = ∠DCB = 27°.

We know that,

An exterior angle is equal to the sum of two opposite interior angles,

Ext. ∠ADC = ∠DBC + ∠DCB = 27° + 27° = 54°.

△ADC is an isosceles triangle with AC = DC,

∴ ∠DAC = ∠ADC = 54°.

Since, sum of angles of triangle = 180°.

Considering △ADC we get,

⇒ ∠DAC + ∠ADC + ∠DCA = 180°

⇒ 54° + 54° + x = 180°

⇒ x + 108° = 180°

⇒ x = 180° - 108° = 72°.

Hence, x = 72°.

In the following figure, find the value of x:

Answer

△ABD is an isosceles triangle with BD = AD,

∴ ∠DBA = ∠BAD = 48°.

We know that,

An exterior angle is equal to the sum of two opposite interior angles,

Ext. ∠ADC = ∠BAD + ∠DBA = 48° + 48° = 96°.

△ADC is an isosceles triangle with AD = DC,

∴ ∠DAC = ∠DCA = x°.

Since, sum of angles of triangle = 180°.

Considering △ADC we get,

⇒ ∠DAC + ∠ADC + ∠DCA = 180°

⇒ x° + 96° + x° = 180°

⇒ 2x° + 96° = 180°

⇒ 2x° = 180° - 96° = 84°

⇒ x° = 42°

Hence, x = 42°.

In the following figure, find the value of x:

Answer

From figure,

∠ACD = 180° - 130° = 50°.

In △ADC, AD = CD.

∴ ∠DAC = ∠ACD = 50°.

∠ADC = 180° - (∠CAD + ∠DCA) = 180° - (50° + 50°) = 80°.

From figure,

∠ADB + ∠ADC = 180°

∠ADB = 180° - ∠ADC = 180° - 80° = 100°.

In △ADB,

∠DBA = ∠DAB = x° (As angles opposite to equal sides are equal)

∠DBA + ∠DAB + ∠ADB = 180°

x° + x° + 100° = 180°

2x° = 80°

x° = 40°

Hence, x = 40°.

In the following figure, find the value of x:

Answer

In △ADC,

DC = CA (Given)

∴ ∠ADC = ∠CAD = a.

∠ADC + ∠DCA + ∠CAD = 180°

⇒ a + 56° + a = 180°

⇒ 2a = 180° - 56°

⇒ 2a = 124°

⇒ a = 62°.

In △ABD,

AD = BD (Given)

∴ ∠ABD = ∠DAB = (x - a)°

In △ABD,

∠ABD + ∠BDA + ∠DAB = 180°

⇒ (x - a)° + (180 - a)° + (x - a)° = 180°

⇒ 2x° - 3a° + 180° = 180°

⇒ 2x° - 3(62°) = 0

⇒ 2x° - 186° = 0

⇒ 2x° = 186°

⇒ x° = 93°

⇒ x = 93.

Hence, x = 93.

In the figure (1) given below, AB = AD, BC = DC. Find ∠ABC.

Answer

Join BD.

In △ABD,

AB = AD

∴ ∠ABD = ∠ADB = a.

∠ABD + ∠ADB + ∠DAB = 180°

a + a + 54 = 180°

2a + 54° = 180°

2a = 126°

a = 63°.

In △BCD,

BC = CD

∴ ∠CDB = ∠DBC = b.

∠CDB + ∠DBC + ∠BCD = 180°

b + b + 116° = 180°

2b + 116° = 180°

2b = 64°

b = 32°.

From figure,

∠ABC = a + b = 63 + 32 = 95°.

Hence, ∠ABC = 95°.

In the figure (2) given below, BC = CD. Find ∠ACB.

Answer

From figure,

∠DAC = 180° - 138° = 42°.

∠BDC = 180° - ∠ADC = 180° - 116° = 64°.

In △BCD,

∠CBD = ∠BDC = 64°.

∠DCB = 180° - (∠CBD + ∠BDC) = 180° - (64° + 64°)

= 180° - 128° = 52°.

In △ADC,

∠DCA = 180° - (∠CAD + ∠ADC) = 180° - (42° + 116°)

= 180° - 158° = 22°.

From figure,

∠ACB = ∠DCB + ∠DCA = 52° + 22° = 74°.

Hence, ∠ACB = 74°.

In the figure (3) given below, AB || CD and CA = CE. If ∠ACE = 74° and ∠BAE = 15°, find the values of x and y.

Answer

From figure,

∠CAE = ∠CEA = a (Angles opposite to equal side are equal.)

In △ACE

⇒ ∠CAE + ∠CEA + ∠ACE = 180°

⇒ a + a + 74° = 180°

⇒ 2a = 106°

⇒ a = 53°.

From figure,

∠CEA + ∠AEB = 180°

a + x = 180°

53° + x = 180°

x = 180° - 53° = 127°.

In △AEB,

⇒ ∠EAB + ∠AEB + ∠ABE = 180°

⇒ 15 + 127° + ∠ABE = 180°

⇒ 142° + ∠ABE = 180°

⇒ ∠ABE = 180° - 142° = 38°.

From figure,

y = ∠ABE (Alternate angles)

∴ y = 38°.

Hence, x = 127° and y = 38°.

In △ABC, AB = AC, ∠A = (5x + 20)° and each of the base angle is $\dfrac{2}{5}$th of ∠A. Find the measure of ∠A.

Answer

Given, each of the base angle is $\dfrac{2}{5}$th of ∠A.

∴ Each base angle = $\dfrac{2}{5}$ × (5x + 20) = (2x + 8)°

In △ABC,

∠A + ∠B + ∠C = 180°

(5x + 20)° + (2x + 8)° + (2x + 8)° = 180°

9x° + 36° = 180°

9x° = 144°

x° = 16°.

(5x + 20)° = 5(16°) + 20° = 80° + 20° = 100°.

Hence, ∠A = 100°.

In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC = CE. Calculate ∠ACE and ∠AEC.

Answer

Since, ABC is an equilateral triangle, ∠A = ∠B = ∠C = 60°.

From figure,

∠ACB + ∠ACE = 180°

60° + ∠ACE = 180°

∠ACE = 120°.

From figure,

∠AEC = ∠CAE (As angles opposite to equal sides are equal)

Let ∠AEC = ∠CAE = y.

From figure,

∠AEC + ∠CAE + ∠ACE = 180°

y + y + 120° = 180°

2y = 60°

y = 30°.

Hence, ∠ACE = 120° and ∠AEC = 30°.

In the figure (2) given below, prove that ∠BAD : ∠ADB = 3 : 1.

Answer

From figure,

∠ADC = ∠CAD (As angles opposite to equal sides of triangle are equal) ......(i)

From figure,

∠BAD = ∠BAC + ∠CAD = ∠BAC + ∠ADC = ∠BAC + ∠ADB .......(ii)

From △ABC,

∠BAC = ∠ACB (As angles opposite to equal sides of triangle are equal) ......(iii)

∠ACB = 180 - (∠ACD) = 180 - [180 - (∠CAD + ∠ADC)] = ∠CAD + ∠ADC ........(iv)

From figure,

∠ADC = ∠ADB

∴ ∠CAD = ∠ADB

From (iv)

∠ACB = ∠CAD + ∠ADC = ∠ADB + ∠ADB = 2∠ADB.

From (iii) we get,

∠BAC = ∠ACB = 2∠ADB.

Substituting value of ∠BAC in (ii) we get,

∠BAD = ∠BAC + ∠ADB = 2∠ADB + ∠ADB = 3∠ADB.

Hence, ∠BAD : ∠ADB = 3∠ADB : ∠ADB = 3 : 1.

Hence, proved that ∠BAD : ∠ADB = 3 : 1.

In the figure (3) given below, AB || CD. Find the values of x, y and z.

Answer

Add Answer fig

From figure,

∠BAD = ∠ADC (Alternate angles)

∴ x = 42°

In △CED,

⇒ 24° + ∠CED + 42° = 180°

⇒ ∠CED + 66° = 180°

⇒ ∠CED = 114°.

From figure,

⇒ ∠CEA + ∠CED = 180°

⇒ ∠CEA + 114° = 180°

⇒ ∠CEA = 66°.

In △CEA,

⇒ y = ∠CEA = 66° (As angles opposite to equal sides are equal)

In △CEA,

⇒ z + y + ∠CEA = 180°

⇒ z + 66° + 66° = 180°

⇒ z + 132° = 180°

⇒ z = 48°.

Hence, x = 42°, y = 66° and z = 48°.

In the adjoining figure, D is the midpoint of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.

Answer

In △BED and △CFD,

∠BED = ∠CFD (Both are equal to 90°)

DE = DF (Given)

BD = DC (As D is the mid-point of BC.)

∴ △BED ≅ △CFD by RHS axiom.

We know that corresponding parts of congruent triangles are equal,

∴ ∠B = ∠C ⇒ AC = AB.

Hence, proved that ABC is an isosceles triangle.

In the adjoining figure, AD, BE and CF are altitudes of △ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.

Answer

AD, BE and CF are altitudes of △ABC and

AD = BE = CF

Considering △BEC and △BFC,

Hypotenuse BC = BC (Common)

Side BE = CF (Given)

As altitudes are perpendicular to the sides,

∠CFB = ∠CEB = 90°

Hence, △BEC ≅ △BFC by RHS axiom.

We know that corresponding parts of congruent triangle are equal,

∴ ∠C = ∠B

⇒ AB = AC (Sides opposite to equal angles) .........(i)

Considering △CFA and △ADC,

AD = CF (Given)

∠ADC = ∠CFA = 90° (As altitudes are perpendicular to sides)

AC = AC (Common)

Hence, △CFA ≅ △ADC by RHS axiom.

We know that corresponding parts of congruent triangle are equal,

∴ ∠A = ∠C

⇒ AB = BC (Sides opposite to equal angles) .........(ii)

From (i) and (ii)

AB = BC = AC

△ABC is an equilateral triangle.

Hence, proved that △ABC is an equilateral triangle.

In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:

(i) △DBC ≅ △ECB

(ii) ∠DCB = ∠EBC

(iii) OB = OC, where O is the point of intersection of BE and CD.

Answer

(i) From figure,

BD = CE (Given)

BC = BC (Common)

∠DBC = ∠ECB (as AB = AC).

∴ △DBC ≅ △ECB by SAS axiom.

Hence, proved △DBC ≅ △ECB.

(ii) We know that corresponding parts of congruent triangle are equal.

∴ ∠DCB = ∠EBC.

Hence, proved that ∠DCB = ∠EBC.

(iii) As △DBC ≅ △ECB,

∠BDO = ∠CEO (By c.p.c.t.)

∠DOB = ∠EOC (Vertically opposite angles)

BD = CE (Given)

Hence, △BOD ≅ △EOC by ASA axiom.

We know that corresponding parts of congruent triangles are equal.

∴ OB = OC.

Hence, proved that OB = OC.

ABC is an isosceles triangle in which AB = AC. P is any point in the interior of △ABC such that ∠ABP = ∠ACP. Prove that

(a) BP = CP

(b) AP bisects ∠BAC.

Answer

Below figure shows the isosceles triangle ABC with the points marked:

(a) Given,

AB = AC

∴ ∠B = ∠C ........(i)

Given, ∠ABP = ∠ACP ........(ii)

Subtracting (ii) from (i) we get,

∠B - ∠ABP = ∠C - ∠ACP

∠PBC = ∠PCB.

∴ BP = CP (As sides opposite to equal angles are equal)

Hence, proved that BP = CP.

(b) We know that,

BP = CP (Proved)

AB = AC (Given)

∠ABP = ∠ACP (Given)

Hence, △ABP ≅ △ACP by SAS axiom.

∠PAB = ∠PAC (Corresponding angles of congruent triangles are equal.)

Thus AP, bisects ∠BAC.

Hence, proved that AP bisects ∠BAC.

ΔPQR is an isosceles triangle such that PQ = QR. If S is a point on QR produced such that PR = RS and ∠QPS = 63°, find ∠PSQ.

Answer

ΔPQR is an isosceles triangle such that PQ = QR.

S is a point on QR produced such that PR = RS.

As we know that in an isosceles triangle, angles opposite to equal sides are equal.

In ΔPQR,

⇒ ∠QPR = ∠QRP = x° (let)....................(1)

In ΔPRS,

⇒ ∠PSR = ∠SPR = y° (let)....................(2)

Given,

⇒ ∠QPS = 63°

From figure,

⇒ ∠QPR + ∠SPR = 63°

⇒ x° + y° = 63° ..................(3)

We know that,

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

The angle ∠PRQ (which is x) is an exterior angle to △PRS at vertex R.

⇒ ∠PRQ = ∠SPR + ∠PSR.

⇒ x° = y° + y°

⇒ x° = 2y°

Substituting the above value of x in equation (1), we get :

⇒ 2y° + y° = 63°

⇒ 3y° = 63°

⇒ y° = $\dfrac{63°}{3}$

⇒ y° = 21°

⇒ ∠PSR = ∠SPR = 21°.

From figure,

⇒ ∠PSQ = ∠PSR = 21°

Hence, ∠PSQ = 21°.

In the adjoining figure, D and E are points on the side BC of △ABC such that BD = EC and AD = AE. Show that △ABD ≅ △ACE.

Answer

Given, AD = AE.

∴ ∠ADE = ∠AED (As angles opposite to equal sides are equal.)

⇒ 180 - ∠ADE = 180 - ∠AED

⇒ ∠ADB = ∠AEC.

BD = EC (Given)

∴ △ABD ≅ △ACE by SAS axiom.

Hence, proved that △ABD ≅ △ACE.

In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that △ADE ≅ △BCE and hence, AEB is an isosceles triangle.

Answer

From figure,

∠ADE = ∠ADC + ∠CDE = 90° + 60° = 150°.

Similarly,

∠BCE = ∠BCD + ∠DCE = 90° + 60° = 150°.

⇒ ∠ADE = ∠BCE.

AD = BC (As sides of squares are equal)

DE = EC (As CDE is an equilateral triangle.)

∴ △ADE ≅ △BCE by SAS axiom.

We know that corresponding parts of congruent triangles are equal.

∴ AE = BE.

Hence, proved that AE = BE i.e. AEB is an isosceles triangle.

In the figure (ii) given below, O is the point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that OCD is an isosceles triangle.

Answer

∠OAD = ∠DAB - ∠OAB = 90° - 60° = 30°.

Similarly,

∠OBC = ∠CBA - ∠OBA = 90° - 60° = 30°.

⇒ ∠OAD = ∠OBC.

AD = BC (As sides of squares are equal).

OA = OB (As OAB is equilateral triangle).

∴ △OAD ≅ △OBC by SAS axiom.

We know that corresponding parts of congruent triangles are equal.

∴ OD = OC.

Hence, proved that OD = OC i.e. OCD is an isosceles triangle.

In the adjoining figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

Answer

In △ABC, ∠A = 90° and AB = AC.

⇒ ∠B = ∠C (As angles opposite to equal sides are equal)

We know,

∠A + ∠B + ∠C = 180°

90° + ∠B + ∠B = 180°

2∠B = 90°

∠B = 45° .

As AD is bisector of ∠A, ∠BAD = ∠CAD = 45°.

AD = AD (Common)

∠ACD = ∠ABD = 45°

Hence, △ABD ≅ △ACD by AAS axiom.

We know that corresponding parts of congruent triangles are equal.

∴ BD = CD .........(i)

In △ABD, ∠BAD = ∠ABD (each = 45°)

⇒ AD = BD (As sides opposite to equal angles are equal) ........(ii)

∴ BC = BD + CD

= BD + BD (Using (i))

∴ BC = 2BD

Using (ii),

BC = 2AD.

Hence, proved that BC = 2AD.