Question

prove that the area of an equilateral triangle described on a side of right angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse.

Answer 1

Here ABCD is a square, AEB is an equilateral triangle described on the side of the square and DBF is an equilateral triangle described on diagonal BD of  the square.
We have to Prove: Ar(ΔDBF) / Ar(ΔAEB) = 2 / 1 
Proof:  If two equilateral triangles are similar then all angles are = 60 degrees.

Therefore, by AAA similarity criterion , △DBF ~

 △AEB

Ar(ΔDBF) / Ar(ΔAEB) = DB2 /AB2 -----------1

We know that the ratio of the areas of two similar triangles is equal to
the square of the ratio of their corresponding sides i .e.
But, we have DB = √2AB     {But diagonal of square is √2 times of its side} -----2
Substitute equation 2 in equation 1, we get
Ar(ΔDBF) / Ar(ΔAEB) = (√2AB )2 / AB2   = 2 AB2 / AB2 = 2

∴ Area of equilateral triangle described on one side os square is equal to half the area of the equilateral triangle described on one of its diagonals.    

Answer 2

Answer:

in triangle YAC and XAB

angle X=angle Y

angle B =angle C

by aa similarity

triangle xba similar to triangle yca

xb/yc =ba/ca=xa/ya

by area theorem

area of xba/area of yca=(xb/yc)^2

(a/√2a)^2=1/2=1:2

proved