Question

In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If \(\angle\)A = 90°, then prove that DE² = BD × EC. (Hint: Show that ∆GBD is similar to ∆ CFE. Use GD = FE = DE.)

Answer 1

Proof : 

꠸DEFG is a square. 

∴ DE = EF = GF = GD (i) [Sides of a square] 

\(\angle\)GDE = \(\angle\)DEF = 90° [Angles of a square] 

∴ seg GD ⊥ side BC, seg FE ⊥ side BC (ii) 

In ∆BAC and ∆BDG,

\(\angle\)BAC ≅ \(\angle\)BDG [From (ii), each angle is of measure 90°]