maths

$In△ABC&△PQR$

$⟹AB=PQ$

$∠ABC=∠PQR=90_{∘}$

$BC=RQ$

$∴△ABC≅△PQR$

$⟹AC=PR(C.P.C.T)$

$In△ABR$

$AB_{2}+BR_{2}=AR_{2}$

$⟹AB_{2}+(BC+CR)_{2}=AR_{2}−(1)$

$In△PCR:$

$⟹PQ_{2}+QC_{2}=PC_{2}$

$⟹PQ_{2}+(QR+CR)_{2}=PC_{2}$

$In(1)&(2),$

$∵AB=PQ(given)$

$andBC=QR(given)$

$⟹AR_{2}=PC_{2}(from(1)&(2))$

$AR=PC−(3)$

$In△ABR&△PQC$

$⟹AB=PQ$

$∠ABR=∠PQC=90_{∘}$

$AR=PC−(from(3))$

$∴△ABR≅△PQCbyR.H.S$

Answer verified by Toppr

The given figure shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and $AD=BC$.

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$△AMC≅△BMD$

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