Question

In Fig. AB=PQ, BC=RQ, AB=BQ and PQ=BQ Prove that $△ABR\cong △PQC.$

Answer 1

$Given:AB=PQ=BQ$

$In△ABC\&△PQR$

$⟹AB=PQ$

$\angle ABC=\angle PQR={90}^{\circ }$

$BC=RQ$

$\therefore △ABC\cong △PQR$

$⟹AC=PR(C.P.C.T)$

$In△ABR$

${AB}^2+{BR}^2={AR}^2$

$⟹{AB}^2+{(BC+CR)}^2={AR}^2-\left(1\right)$

$In△PCR:$

$⟹{PQ}^2+{QC}^2={PC}^2$

$⟹{PQ}^2+{(QR+CR)}^2={PC}^2$

$In\left(1\right)\&\left(2\right),$

$\because AB=PQ\left(given\right)$

$andBC=QR\left(given\right)$

$⟹{AR}^2={PC}^2\left(from\right(1\left)\&\right(2\left)\right)$

$AR=PC-\left(3\right)$

$In△ABR\&△PQC$

$⟹AB=PQ$

$\angle ABR=\angle PQC={90}^{\circ }$

$AR=PC-\left(from\right(3\left)\right)$

$\therefore △ABR\cong △PQCbyR.H.S$