Question

In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If ∠A = 90° , then prove that DE² = BD × EC. 

(Hint: Show that ∆GBD is similar to ∆ CFE. Use GD = FE = DE.)

Answer 1

\(\Box \)DEFG is a square. 

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

∠GDE = ∠DEF = 90° [Angles of a square] 

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

In ∆BAC and ∆BDG, 

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

∠ABC ≅ ∠DBG [Common angle] 

∴ ∆BAC – ∆BDG (iii) [AA test of similarity]  

In ∆BAC and ∆FEC, 

∠BAC ≅ ∠FEC [From (ii), each angle is measure 90°] 

∠ACB ≅ ∠ECF [Common angle] 

∴ ∆BAC – ∆FEC (iv) [AA test of similarity] 

∴ ∆BDG – ∆FEC [From (iii) and (iv)] 

∴ BD/EF = GD/EC  (v) [Corresponding sides of similar triangles] 

∴ BD/EF = GD/EC  [From (i) and (v)] 

∴ DE² = BD × EC