Question

In ∆ABC, AB = AC. Side BC is produced to D. Prove that AD² − AC² = BD. CD

Answer 1

Draw AE⊥BC, meeting BC at D. 

Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle. 

So, BE = CE 

And DE+CE=DE+BE=BD

AD² = AE² + DE² 

⇒ AE² = AD² − DE² …(i) 

In ∆ACE,