# Triangles – Exercise 7.2 – Class IX

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 : (i) OB = OC (ii) AO bisects ∠ A

Solution:

AB = AC such that ∠ABC  =  ∠ACB [angles opposite to equal sides are equal ]

1/2∠ABC  =  1/2∠ACB

⇒ ∠CBO = ∠BCO  [OB and OC are bisectors of ∠B and ∠C respectively]

⇒ OB  = OC [sides opposite to  equal angles are equal]

Again, 1/2∠ABC  =  1/2∠ACB

⇒ ∠ABO = ∠ACO [ OB  and OC are bisectors of ∠B and  ∠C respectively]

In ∆ABO  and ∆ACO,  we have

AB = AC [given]

OB = OC [proved above]

∠ABO = ∠ACO [proved above]

∆ABO ≅ ∆ACO  [SAS congruence]

⇒ ∠ABO = ∠CAO [CPCT]

⇒ AO bisects ∠A

1. In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ ABC is an isosceles triangle in which AB = AC.

Solution:

In ∆ABD and ∆ACD , we have

BD = CD [ because AD bisects  BC]

∆ABD ≅ ∆ACD [SAS congruence]

AB = AC [CPCT]

Hence, ∆ABC is an isosceles triangle.

1. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Solution:

In ∆ABC,

given, AB = AC

⇒ ∠B = ∠C [angles opposite to equal sides are equal in a triangle]

Now, in ∆BFC and ∆CEB,

∠BFC = ∠CEB = 90˚

∠FBC = ∠ECB [which is already proven]

BC = BC [Common]

∆BFC = ∆CEB [AAS congruence]

Hence, BE = CF [CPCT]

1. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Solution:

(i) In ∆ABE and ∆ACF, we have,

Given, BE = CF

∠BAE = ∠CAF [common]

∠BEA = ∠CFA = 90˚

So, ∆ABE ≅ ∆ACF [AAS congruence]

(ii)AB = AC [CPCT]

That is, ∆ABC is an isosceles triangle.

1. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD = ∠ ACD.

Solution:

∆ABC is an isosceles triangle

AB = AC …………(1)

Also, ∆DBC is isosceles triangle

DB = DC …………(2)

To prove: ∆ABD≅∆ACD

Proof:

In ∆ABC and ∆DBC , we have,

AB = AC from (1)

BD = DC from(2)

⇒∠DBC = ∠DCB ……(2) (equal sides have equal angles opposite to them)

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

∠ABC = ∠DBC = ∠ACB + ∠DCB

∆ABD ≅ ∆ACD (SSS congruence rule)

1. ΔABC is an isosceles triangle in which AB = AC Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.

Solution:

Given in ΔABC, AB = AC
⇒ ∠ABC = ∠ACB (Since angles opposite to equal sides are equal)
Also given that AD = AB
⇒ ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)

∠ACB + ∠ACD = ∠ABC + ∠ADC

∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD (since AB = AC = AD)

(∠CBD = ∠ABC , ∠CDB = ∠ADC)
In ΔBCD, ∠CBD + ∠BCD + ∠CDB = 180°

∠BCD + ∠BCD = 180°

2∠BCD = 180°

∠BCD = 180°/2 = 90°
Thus BCD is a right angled triangle

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

Solution: If AB = AC then angles opposite to these sides will be equal.

∠BAC + ∠ABC + ∠ACB = 180°

90° + ∠ABC + ∠ACB = 180°

∠ABC + ∠ACB = 180° –  90°

∠ABC + ∠ACB = 90°

∠ABC = ∠ACB = 45°

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

Solution: In ∆ABC,

AB = BC = CA, (the ∆ ABC is an equilateral triangle)

∠BAC = ∠ABC = ∠ACB (Since, AB = BC = CA)

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + ∠BAC + ∠BAC = 180°

3∠BAC = 180°

∠BAC = 180°/3 = 60°

∠BAC = ∠ABC = ∠ACB = 60°