In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°.

Given: ▢DEFG is a square.

To prove: DE2 = BD × EC

Proof:

Given that ▢DEFG is a square, therefore, all its sides are equal.

Thus, DE = EF = FG = GD (from the definition of a square) …(1)

Also, the angles of the square are 90 degrees, so ∠GDE = ∠DEF = 90°.

Since the diagonals of a square bisect each other and are perpendicular to each other, we have seg GD ⊥ sides BC and seg EF ⊥ side BC.

Consider ΔBAC and ΔBDG,

∠BAC and ∠BDG are right angles (as they lie in a square)

∠ABC and ∠DBG are equal (as they are opposite angles of parallelogram BDGF)

Thus, by AA test of similarity, we get

ΔBAC ~ ΔBDG …(2)

Similarly, ΔBAC ~ ΔFEC …(3)

From equations (1), (2), and (3), we can conclude that

ΔBDG ~ ΔFEC

Therefore, corresponding sides of the triangles are in proportion.

Hence, "BD"/"EF" = "GD"/"EC"

Using equation (1), we get "BD"/"DE" = "DE"/"EC"

Multiplying both sides by DE, we get

DE² = BD × EC

Hence, proved.