Question

In the given figure, 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 is ____.

• A. $AD=BF$
• B. $AD>BF$
• C. $AD<BF$
• D. None of these

Answer 1

The correct option is A

$AD=BF$

In $\Delta ACDand\Delta FCB$, we have:
$\angle ACD={90}^{\circ }+\angle BCA$ and
$\angle FCB={90}^{\circ }+\angle BCA$
CF = CA; ($\because$ Sides of a square)
CD = CB ($\because$ Sides of a square)
$\therefore \Delta ACD\cong \Delta FCB$ [by S.A.S. axiom]
$\Rightarrow AD=BF$ (CPCT)