Question

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,

• A. $AD=BF$
• B. $AD>BF$
• C. $AD<BF$
• D. lengths of sides AD and BF cannot be compared.

Answer 1

The correct option is A

$AD=BF$

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