In the adjoining figure, ABC is a right angled triangle at B. ADEC and BCFG are squares. Prove that AF = BE.

Given, ADEC and BCFG are squares.

Considering △BCE and FCA we get,

From figure,

∠BCE = ∠BCA + 90

∠ACF = ∠BCA + 90

∠BCE = ∠ACF

AC = CE (Sides of squares are equal)

BC = CF (Sides of squares are equal)

△BCE ≅ △FCA (By SAS axiom).

We know that corresponding parts of congruent triangles are equal.

∴ AF = BE.

Hence, proved that AF = BE.