ABC is a triangle, right angled at B. If BCDE is a square on the side BC and ACFG is a square on AC, then AD = .

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Solution

The correct option is C BF
In $Δ A C D a n d Δ B C F$ , we have

$∠ A C D = 90^{\circ} + ∠ B C A$ and

$∠ B C F = 90^{\circ} + ∠ B C A$

CF = CA (Sides of a square)

CD = CB (Sides of a square)

$∴ Δ D C A ≅ Δ B C F$
(by SAS congruence rule)

$\Rightarrow A D = B F$ (CPCT)

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