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Given: ∠ABC = 90° , BCDE is a square on side BC and ACFG is a square on AC

To prove: AD = EF

Proof:

Since BCDE is square,

∠BCD = 90° …(1)

In ∆ACD,

∠ACD = ∠ACB + ∠BCD

= ∠ACB + 90° …(2)

In ∆BCF,

∠BCF = ∠BCA + ∠ACF

Since ACFG is square,

∠ACF = 90° …(3)

From 2 and 3, we have,

∠ACD = ∠BCF ….(4)

Thus in ∆ACD and ∆BCF, we have,

AC = CF ...sides of square

∠ACD = ∠BCF …from 4

CD = BC … sides of square

Thus by SAS property of congruence,

∆ACD ≅ ∆BCF

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ AD = BF