Question

In the figure given below triangles ABC is right - angled at B . ABPQ and ACRS are square . Prove that :
CQ = BS

Answer 1

Given ; A $\Delta ABC$ is right-angled at $B.$
$ABPQ$ and $PQRS$ are square
We need to prove that
$\Delta ACQ\cong \Delta ASB$
$\angle QAB={90}^0$
$\angle SAC={90}^o$
$\angle QAB=\angle SAC$
Adding $\angle BAC$ to both sides of (3) , we have
$\angle QAB+\angle BAC=\angle SAC+\angle BAC$

$\Rightarrow \angle QAC=\angle SAB$ ...(1)

Now, In $\Delta ACQ$ and $\Delta ASB$,
$QA=QB$ ...[Sides of a square ABPQ]
$\angle QAC=\angle SAB$ ...[From (1)]
$AC=AS$ ....[Sides of a squre ACRS]
So, By angle-angle-side criterion of congruence,
$\Delta ACQ\cong \Delta ASB$

The corresponding parts of the congruent triangles are congruent
$CQ=BS$ [c.p.c. t]