Proof: 1 Mark
Steps: 2 Marks
Consider two triangles, $\mathrm{Δ}ABC$ and $\mathrm{Δ}PQR$ in which,

$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$
$\mathrm{âˆ}ACB=\mathrm{âˆ}PRQ$
$AB=PQ$
We know that, 
$\mathrm{âˆ}ABC+\mathrm{âˆ}ACB+\mathrm{âˆ}BAC={180}^0$
$\mathrm{âˆ}BAC={180}^0−(\mathrm{âˆ}ABC+\mathrm{âˆ}ACB)....\left(i\right)$
Similarly,
$\mathrm{âˆ}QPR={180}^0−(\mathrm{âˆ}PQR+\mathrm{âˆ}PRQ)......\left(ii\right)$
From (i) and (ii), 
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$
Now, In  $\mathrm{Δ}ABC$ and $\mathrm{Δ}PQR$
$\mathrm{âˆ}ABC=\mathrm{âˆ}PQR$  [Given]
$\mathrm{âˆ}BAC=\mathrm{âˆ}QPR$   [ Proved above]
$AB=PQ$  [Given]
$⇒\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}PQR$  [ ASA congruency rule]
$⇒$ These triangles are always congruent.