Question

Prove that the area of  Δ BCE described on one side BC of a square ABCD is one half the area of the similar Δ ACF described on the diagonal AC. 

Answer 1

In right angled triangle ABC ,

By Pythagoras Theorem , AB² + BC² = AC²

Given , Δ BCE ∼ Δ ACF

`("Ar" triangle "BCE")/("Ar" triangle "ACF") = "BC"^2/"AC"^2`

[The ration of areas of two similar triangle is equal to the ratio of square of their corresponding sides.]

`= "BC"^2/"AC"^2`

`= 1/2`

Required ratio is 1 : 2.