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  • In the given figure, ABC is a triangle, right angled at B. If

In the given figure,

ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC. Then

Option: 1

A D\neq F B


Option: 2

A D=F B


Option: 3

A D=\sqrt2F B


Option: 4

None of the above


Answers (1)

best_answer

\begin{align*} &\text{In } \triangle ACD \text{ and } \triangle FCB, \text{ we have:} \\ &\angle ACD = 90^{\circ} + \angle BCA \\ &\angle FCB = 90^{\circ} + \angle BCA \\ \Rightarrow &\angle ACD = \angle FCB \\ &CA = CF \quad \text{(sides of the same square)} \\ &CD = CB \quad \text{(sides of the same square)} \\ \therefore &\triangle ACD \cong \triangle FCB \quad \text{(by SAS-criteria)} \\ \text{Hence, } &AD = FB \quad \text{(c.p.c.t.)} \end{align*}

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Deependra Verma

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