Q40 of 97 Page 187

In the given figure, ABC is a triangle in which AB=AC. If D be a point on BC produced, prove that AD>AC.

Given: AB=AC


To prove: AD>AC


Proof:


In ∆ABC,


∠ACD = ∠B + ∠BAC


= ∠ACB + ∠BAC …as ∠C = ∠B as AB = AC


= ∠CAD + ∠CDA +∠BAC …as ∠ACB = ∠CAD + ∠CDA


∴∠ACD > ∠CDA


So the side opposite to ∠ACD is the longest


∴AD > AC


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