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Question
ΔABC is isosceles with AB = AC. If BC is extended to D, then prove that AD > AB.
Solution
Using the exterior angle and property in ΔACD, we have
∠ACB = ∠CDA + ∠CAD
⇒ ∠ACB > ∠CDA ------(1)
Now, AB = AC
Now, ∠ACB = ∠ABC ------(2)
From (1) and (2)
∠ABC > ∠CDA
It is known that, in a triangle, the greater angle has the longer side opposite to it.
Now, In ΔABD, er have ∠ABC > ∠CDA
∴ AD > AB.
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