In the given figure, ABC is a triangle, right-angled at B. BCDE is a square on the side BC and ACFG is a square on AC. Then,
AD=BF
In ΔACD and ΔFCB, we
have,
∠ACD=90∘+∠BCA and
∠FCB=90∘+∠BCA
⟹∠ACD=∠FCB
CF = CA; (∵ Sides of square ACFG)
CD = CB (∵ Sides of a square BCDE)
∴ΔACD≅ΔFCB
[by S.A.S. axiom]
⇒AD=BF (CPCT)