Given : `triangleABC ~triangle DEF AL bot BC and DM bot EF`
To prove : `(ar(triangleABC))/(ar(triangleDEF))= (AL^(2))/(DM^(2))`
proof: in ` triangleALB and triangleDME`,
` angleB= angleE ( given triangleABC ~ triangleDEF)`
`angleALB = angleDME= 90^(@)` (each)
` triangleABL ~ triangleDEM` (by AA similarity)
` Rightarrow" " (AB) /(DE) = (AL)/(DM) ..(1) `
Now , `(ar(triangleABC))/(ar(triangleDEF))= (AB^(2))/(DE^(2)) ...(2)`
( the ratio of areas of two similar triangles is equal to ratio of squares of the corresponding sides)
Form (1) and (2) we have
`(ar(triangleABC))/(ar(triangleDEF))= (AL^(2))/(DM^(2))` Hence proved.
