Given: A square ABCD and equilateral `triangleBCE and triangleACF` have been described on side BC and diagonal AC respectively.
To prove: `ar(triangleBCE) 1/2 ar(triangleACF)`
Proof: Since each of the `triangleBCE and triangleACF` is an equilateral triangles so each angle of each are of them is `60^(@)`.
Hence ` triangleBCE ~ triangleACF`
`(ar(triangleBCE))/(ar(triangleACF))= (BC^(2))/(AC^(2))= (BC^(2))/(2(BC)^(2))=1/2`
Hence, `ar(triangleBCE)1/2 xx ar(triangleACF)`