91Ó°ÊÓ

It is given that,$∠4â‰\dots ∠7$and the angle adjacent to congruent sides are congruent, that is, width="68">∠5≅∠6.

If two angles of triangle are congruent then the sides opposite to those angles are congruent.

As $∠5â‰\dots ∠6$. Therefore, by converse of isosceles triangle theorem, $\stackrel{¯}{AL}â‰\dots \stackrel{¯}{AM}$.

If two angles and included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

Here, $∠4â‰\dots ∠7$,$\stackrel{¯}{AL}â‰\dots \stackrel{¯}{AM}$ and $∠1â‰\dots ∠3$, then by ASA postulate $\mathrm{Δ}ABLâ‰\dots \mathrm{Δ}ACM$.

As $\mathrm{Δ}ABLâ‰\dots \mathrm{Δ}ACM$, therefore, by congruent parts of congruent triangles, $\stackrel{¯}{AB}â‰\dots \stackrel{¯}{AC}$.

As,$\stackrel{¯}{AB}â‰\dots \stackrel{¯}{AC}$, then by definition of isosceles triangle,$\mathrm{Δ}ABC$ is an isosceles triangle.

Hence it is proved that, $\mathrm{Δ}ABC$is an isosceles triangle.