Given ∆ABC is an isosceles triangle in which ∠B is right angled i.e. 90°
⇒ AB = BC
In right angled ∆ABC, by Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2
⇒ (AB)2 + (AB)2 = (AC)2
[∵ABC is an isosceles triangle, AB =BC]
⇒ 2(AB)2 = (AC)2
⇒ (AC)2 = 2(AB)2 …(i)
It is also given that ∆ABE ~ ∆ADC
And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.
∴ ar(∆ABE) : ar(∆ADC) = 1 : 2