In figure, the vertices, of square DEFG are on the sides of `DeltaABC`. `/_A=90^(@)`. Then prove that `DE^(2)=BDxxEC`.

Explanation

`squareDEFG` is a square (Given)
`:.DE=EF=GF=GD`
………….(Sides of a square )………….1
`/_GDE=/_DEF=90^(@)`………….(Angles of a square)
`:.` seg `GD_|_` sides BC and seg `EF_|_` side BC
In `DeltaBAC` and `DeltaBDG`,
`/_BAC~=/_BDG`….......(Each measures `90^(@)`)
`/_ABC~=/_DB` (Common angles)
`:.DeltaBAC~DeltaBDG` ...........(AA test of similarity) .......2
Similarly `DeltaBAC~DeltaFEC`.......3
`:.DeltaBDG~Delta FEC` ...........[From 2 and 3]
`:.(BD)/(EF)=(GD)/(EC)`.............(Corresponding sides of similar triangles are in proportion)
`:.(BD)/(DE)=(DE)/(EC)`............[From 1]
`:.DE^(2)=BDxxEC`.