**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

**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

∠ADB = ∠ADC = 90˚

BD = CD [ because AD bisects BC]

AD = AD [common]

∆ABD ≅ ∆ACD [SAS congruence]

*⸫* AB = AC [CPCT]

Hence, ∆ABC is an isosceles triangle.

**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]

**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.

**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)

AD = AD (common)

⇒∠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)

**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

∠BCD = ∠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

**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°

**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°

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