### Solution: General enunciation: ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.

**Particular enunciation:** Given that, ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. It is required to prove that AD>AB.

**Proof:** In **∆**ABC AB = AC

∴∠ABC = ∠ACB

*Now ∠ACD be an exterior angle of ∆ABC*

∴∠ACD >interior opposite ∠ABC

⟹∠ACD >∠ACB [∵∠ABC = ∠ACB]

Again ∠ACB is the external angle of **∆**ACD.

∴∠ACB>∠ADC

Since ∠ACD >∠ACB and ∠ACB>∠ADC

∴∠ACD >∠ADC

Now In **∆**ACD

∠ACD >∠ADC

⟹AD>AC

⟹AD>AB [AB = AC]

∴ AD>AB **(Proved)**

## Problem: In the **∆**ABC, AB = AC and D is any point on BC. Prove that AB>AD.

**Solution: General equation:** In the **∆**ABC, AB = AC and D is any point on BC. Prove that AB>AD.

**Particular Enunciation:** Given that, in the **∆**ABC, AB = AC and D is any point on BC. It is required to prove that AB>AD.

**Proof:** Given that in **∆**ABC, AB = AC

∴∠ABC = ∠ACB [∵ the angles opposite to equal sides are equal]

Again, ∠ADB is an exterior angle of **∆**ACD

∴∠ADB>interior opposite angle ∠ACD

⟹∠ADB>∠ACB

⟹∠ADB>∠ABC [∵∠ABC = ∠ACB]

⟹∠ADB>∠ABD

*Now in triangle ∆ABD,*

∠ADB>∠ABD [∵ the angles opposite to equal sides are equal]

⟹AB>AD (Proved)

#### Problem: In the **∆**ABC, AB ⊥ AC and D is any point on AC. Prove that BC>BD.

**Solution: General enunciation:** In the **∆**ABC, AB ⊥ AC and D is any point on AC. Prove that BC>BD.

**Particular enunciation:** Given that, in the **∆**ABC, AB ⊥ AC and D is any point on AC. It is required to prove that BC>BD.

**Proof:** Given that, in **∆**ABC, AB ⊥ AC

∠A = 90^{0}

∴ BC be the hypotenuse.

∴ ∠BAC > ∠BAC

Again ∠BDC is an exterior angle of **∆**ABD.

∴∠BDC>interior opposite ∠BAD

⟹∠BDC>∠BAC

⟹∠BDC>∠BCA [∵∠BAC > ∠BAC]

⟹∠BDC>∠BCD

Now, in **∆**BCD

∠BDC>∠BCD

⟹BC>BD [∵ the angles opposite to equal sides are equal]

∴ BC>BD *(Proved)*