ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ΔABE and ΔACD.
Given ΔABC is an isosceles triangle in which ∠B = 90°
⇒ AB = BC
By Pythagoras theorem, we have
⇒[Since AB = BC]
∴ → (1)
It is also given that ΔABE ~ ΔACD
Recall that ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.
∴ar(ΔABE):ar(ΔACD)=1:2