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 ΔABEandΔACD.
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Solution
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