## MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

- OB = OC
- AO bisect ∠A.

Solution:

Given

AB = AC

∠1 = ∠2, ∠3 = ∠4

To prove:

- OB = OC
- ∠5 = ∠6

Proof:

In ∆ABC,

AB = AC

∠B = ∠C

\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C

∠1 = ∠3 or ∠2 = ∠4

In ∆OBC

∠2 = ∠4

and so OB = OC

(In a A, sides opposite to equal angles are equal)

In ∆ABO and ∆ACO

BO = CO (proved)

∠1 = ∠3 (proved)

AB = AC (given)

∆ABO = ∆ACO (by SAS)

and so ∠5 = ∠6 (by CPCT)

Question 2.

In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:

Given

∠ABC = ∠ADC

To prove:

AB = AC

Proof:

In A ABD and A ACD

BD = CD (given)

∠ADB = ∠ADC (given each 90°)

AD = AD (common)

∴ ∆ABD = ∆ACD (BySAS)

and so AB = AC (by CPCT)

Question 3.

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:

Given

AB = AC

∠E = ∠F (each 90°)

To prove: BE = CF

Proof:

In ∆ABE and ∆ACE

∠A = ∠A (common)

∠E = ∠F (each 90°)

AB = AC (given)

∆ABE = ∆ACE (byAAS)

and so BE = CF (by CPCT)

Question 4.

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

- ∆ABE ≅ ∆ACF
- AB = AC,

i. e., ABC is an isosceles triangle.

Solution:

Given

BE = CF

∠E = ∠F (each 90°)

To prove:

- ∆ABE = ∆ACE
- AB = AC

Proof:

In ∆ABE and ∆ACF

∠A = ∠A (common)

BE = CF (given)

∠E = ∠F (each 90°)

∆ABE = ∆ACF (by AAS)

and so AB = AC (by CPCT)

Question 5.

ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:

Given

AB = AC

BD = CD

To prove

∠ABD = ∠ACD

Construction: Join AD

Proof:

In ∆ABD and ∆ACD

AD = AD (common)

AB = AC (given)

BD = CD (given)

∆ABD = ∆ACD (by SSS)

and so ∠ABD = ∠ACD (by CPCT)

Question 6.

∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.

Solution:

Given: AB = AC

AD = AB

To show: ∠BCD = 90°

i.e., ∠2 + ∠3 = 90°

Proof:

AB = AC …..(1)

AB = AD …..(2)

From (1) and (2), we get

AC = AD

In ∆ABC

AB = AC

∠1 = ∠2

(In a A, angles opposite to equal sides are always equal) …p) …(3)

In ∆ACD

AC = AD

∠3 = ∠4

(In a A, angles opposite to equal sides are always equal) …(4)

In ∆BCD

∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)

∠2 + ∠2 + ∠3 + ∠3 = 180°

(∴ ∠1 = ∠2, ∠3 = ∠4)

2 (∠2 + ∠3) = 180°

(∠2 + ∠3) = 90°

∠BCD = 90°

Question 7.

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

Solution:

In ∆BAC

AB =AC

∠B = ∠C = x

∠A + ∠B + ∠C= 180°

∠B + ∠C = 180° – 90°

∠B + ∠C = 90°

2x = 90°

x = \(\frac{90^{\circ}}{2}\)

Question 8.

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

Solution:

Given

ABC is an equilateral ∆

i. e., AB = BC = AC

To prove

∠A = ∠B = ∠C = 60°

Proof:

In ∆BAC

AB = AC

∠B = ∠C

(In a A, angles opposite to equal sides are always equal) ……(1)

AC = BC

∠A = ∠B

(In a A, angles opposite to equal sides are always equal) …..(2)

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

∠A = ∠B = ∠C = x (say)

∠A + ∠B + ∠C = 180° (ASP)

⇒ x + x + x = 180°

⇒ 3x = 180°

⇒ x = \(\frac{180^{\circ}}{3}\) = 60°

∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.

SSS (Side-Side-Side) Congruence Theorem:

Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

Given

In ∆s ABC and DEF we have,

AB =DE

BC = EE

and AC = DF

To prove:

∆ABC = ∆DEF

Construction:

Suppose BC is the longest side.

Draw EF such that EE = AB and FEG = ∠CBA.

Join GF and DG.

Proof:

In ∆s ABC and GEE, we have

AB = GE (Const.)

∠ABC = ∠GEF (Const.)

and BC = EF (Given)

∴ ∆ABC = ∆GEF (SAS Cong. Axiom)

∠A = ∠G (CPCT) …..(1)

and AC = GF (CPCT) …..(2)

Now AB = EG (Const.)

AB = DE (Given)

∴ DE = EG ……(3)

Similarly, DF = GF ……(4)

In ∆EDG

DE = EG (Proved above)

∴ ∠A = ∠2 (∠s opp. equal side) …..(5)

In ∆DFG

FD = FG (Proved above)

∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)

∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]

i. e. ∠D = ∠G …..(7)

But ∠G = ∠A [From (1)]

∴ ∠A = ∠D …..(8)

In ∆s ABC and DEF,

AB – DE (Given)

AC = DF (Given)

∠A = ∠D [From (8)]

∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.

RHS (Right Angle Hypotenuse Side) Congruence Theorem:

Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.

Given

In ∆s ABC and DEF,

∠B = ∠E = 90°

AC = DF

BC = EF.

∆ABC ≅ ∆DEF

Construction:

Produce DE to M so that

EM = AB, Join ME.

Proof:

In ∆s ABC and MEF

AB = ME (Const.)

BC = EE (Given)

∠B = ∠MEF (each 90°)

∴ ∆ABC = ∆MEF (SAS Cong. Axiom)

Hence ∠A = ∠M (CPCT) …(1)

AC = MF (CPCT) …(2)

Also AC =DF (Given)

∴ DF = MF

∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)

From (1) and (3), we have

∠A = ∠D …..(4)

Now, in ∆s ABC and DEF, we have

∠A = ∠D [From (4)]

∠B = ∠E (Given)

∴ ∠C = ∠F …..(5)

Again, in ∆s ABC and DEF, we have

BC = EF (Given)

AC = DF (Given)

∠C = ∠F [From (5)]

∴ ∆ABC = ∆DEF (SAS Cong. Axiom)