Question

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

Answer 1

Given: ${\displaystyle \square }$DEFG is a square.

To prove: DE² = BD × EC

Proof:  DEFG is a square

In ΔGBD and ΔAGF,

∠GDB = ∠FAG = 90°   ...(Each measures 90°)

∠AGF = ∠DBG    ...(Common angle)

ΔGBD ∼ ΔAGF      ...(AA Similarity Test) (1)

In ΔCFE and ΔAGF,

∠FAG = ∠CEF =  90°    ...(Each measures 90°)

∠AFG = ∠ECF       ...(Common angle)

ΔCFE ∼ ΔAGF      ...(AA Similarity Test) (2)

From (1) and (2), we get

ΔCFE ∼ ΔGBD   

∴ ${\displaystyle \frac{\text{CE}}{\text{GD}}=\frac{\text{FE}}{\text{BD}}}$  ...(Corresponding sides of similar triangles are in proportional)

∴ ${\displaystyle \frac{\text{CE}}{\text{DE}}=\frac{\text{DE}}{\text{BD}}}$  ...(∵ GD = FE = DE)

∴ DE² = BD × CE

Hence proved.

Answer 2

Given: ${\displaystyle \square }$DEFG is a square.

To prove: DE² = BD × EC

Proof: ${\displaystyle \square }$DEFG is a square.      ...(Given)

∴ DE = EF = GF = GD   ...(Sides of a square) (I)

∠GDE = ∠DEF = 90°    ...(Angles of a square)

∴ seg GD ⊥ sides BC and seg EF⊥ side BC.

In ΔBAC and ΔBDG,

∠BAC ≅ ∠BDG      ...(Each measures 90°)

∠ABC ≅ ∠DBG     ...(Common angles)

∴ ΔBAC ~ ΔBDG     ...(AA test of similarity) (II)

Similarly,

ΔBAC ~ ΔFEC      ...(III)

∴ ΔBDG ~ ΔFEC    ...[From II and III]

∴ ${\displaystyle \frac{\text{BD}}{\text{EF}}=\frac{\text{GD}}{\text{EC}}}$   ...(Corresponding sides of similar triangles are in proportion)

∴ ${\displaystyle \frac{\text{BD}}{\text{DE}}=\frac{\text{DE}}{\text{EC}}}$  ...(From 1)

∴ DE² = BD × EC