Question

In the figure, given below, triangle ABC is right-angled at B: ABPQ and ACRS are squares.Prove that :

(i) $\Delta$ ACQ and $\Delta$ ASB are congruent.

(ii) CQ =BS.

Answer 1

$\angle QAC$ = $\angle QAB+\angle BAC={90}^0+x$
$\angle SAB=\angle CAS+\angle BAC={90}^0+x$

THUS, $\angle QAC=\angle SAB$

In $△QAC$ and $△$ BAS
QA = AB (sides of same square)
AS = AC (sides of same square)
$\angle QAC=\angle SAB$ Proved

Thus, $\Delta$ ACQ is congruent to $\Delta$ ASB (SAS)

CQ=BS (cpctc)