Question

In the figure given along side, squares ABDE and AFGC are drawn on the side AB and the hypotenuse AC of the right triangle ABC. If BH is perpendicular to FG then $\Delta EAC\cong \Delta BAF$.
If true then enter $1$ and if false then enter $0$ :

Answer 1

In the given figure, squares $ABDE$ and $AFGC$ are drawn on the side $AB$ and hypotenuse $AC$ of the right angled triangle $ABC$.

In $△EAC,$

$\angle EAC=\angle EAB+\angle BAC$

$⟹\angle EAC={90}^o+\angle BAC$ ....... $\left(i\right)$ $[\angle EAB$ is the angle of square $ABDE]$

In $△BAF,$

$\angle BAF=\angle CAF+\angle BAC$

$⟹\angle BAF={90}^o+\angle BAC$ ....... $\left(ii\right)$ $[\angle CAF$ is the angle of square $ACGF]$

From $\left(i\right)$ and $\left(ii\right)$

$\angle EAC=\angle BAF$ ....... $\left(iii\right)$

Now, in $△EAC$ and $△BAF$

$EA=BA$ ..... [Sides of a square $EABD$]

$\angle EAC=\angle BAF$ ............. From $\left(iii\right)$

$AC=AF$ ......... [Sides of a square $ACGF$]

$\therefore △EAC\cong BAF$ ....... [By SAS rule of congruence]

Thus, the statement is true.

Hence, the answer is $1$.