If two triangles are similar; prove that the ratio of the corresponding sides is same as the corresponding angle bisector segments.

Explanation

Given : `triangleABC ~ triangleDEF` in which AX and DY are the bisectors of `angleA and angleD` respectively.
To prove: `(ar(triangleABC))/(ar(triangleDEF)) = (AB^(2))/(DE^(2))`
Proof: Given `triangleABC ~ triangleDEF`
Therefore, `angleA= angleD`
`Rightarrow"" 1/2 angleA= 1/2 angleD`
Now, in `triangleABX and triangleDEY` , we have
` angleBAX = angleEDY`
and ` angleB= angleE`
`triangleABX~ triangleDEY` (by AA similartiy)
Hence ` (AB)/(DE)=(AX)/(DY)` ...(1) (given)
Now ,in `triangleABC and triangleDEF`
`triangleABC ~ triangleDEF`
therefore, `(ar(triangleABC))/(ar(triangleDEF))= (AB^(2))/(DE^(2))` ....(2)
( in similar triangles the ratio of areas of triangles is equal to ratio of squares of their corresponding sides)
form (1) and (2) , we have
`(ar(triangleABC))/(ar(triangleDEF))= (AX^(2))/(DY^(2))`