Question

In $\Delta ABCand\Delta PQR$, AB = PQ, BC = QR; CB and RQ are extended to X and Y respectively. $\angle ABX=\angle PQY$. Prove that $\Delta ABC=\Delta PQR$.

Answer 1

In $\Delta ABC$ and $\Delta PQR$

$AB=PQ$

And $BC=QR$

$CB$ And $RQ$ extended to $X$ and $Y$

$\angle ABX=\angle PQY$

$\angle ABC={180}^o-\angle ABX={180}^o\angle PQY$

$\angle PQR={180}^o-\angle PQY=\angle ABC$

We get in both triangles

$AB=PQ$

$BC=QR$

And $\angle ABC=\angle PQR$

Then both triangle are congruent

i.e. $\Delta ABC=\Delta PQR$