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Home / Geometry / ABC is an isosceles triangle and AB = AC. The side BC is extended up to D. Prove that AD>AB.

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

### 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.

Now In ACD

## 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

Now in triangle ABD,

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

#### 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 = 900

∴ BC be the hypotenuse.

∴ ∠BAC > ∠BAC

Again ∠BDC is an exterior angle of ABD.

⟹∠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)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...