Question

Given:$\mathbf{\angle }\mathbf{4}\mathbf{\cong }\mathbf{\angle }\mathbf{7}\mathbf{;}\mathbf{\angle }\mathbf{1}\mathbf{\cong }\mathbf{\angle }\mathbf{3}$

Prove:$\mathrm{\Delta }\mathbf{ABC}$ is isosceles.

Short answer

It is proved that $\Delta ABC$is an isosceles triangle.

Step-by-step solution

– Description of step.

It is given that,$\angle 4\cong \angle 7$ and the angle adjacent to congruent sides are congruent, that is, $\angle 5\cong \angle 6$.

– Converse of isosceles triangle theorem.

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

As $\angle 5\cong \angle 6$. Therefore, by converse of isosceles triangle theorem, $\overline{AL}\cong \overline{AM}$.

– ASA postulate.

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, $\angle 4\cong \angle 7,\overline{AL}\cong \overline{AM}$and $\angle 1\cong \angle 3$, then by ASA postulate $\Delta ABL\cong \Delta ACM$.

– Description of step.

As $\Delta ABL\cong \Delta ACM$, therefore, by congruent parts of congruent triangles, $\overline{AB}\cong \overline{AC}$.

– Description of step.

As,$\overline{AB}\cong \overline{AC}$, then by definition of isosceles triangle, $\Delta ABC$is an isosceles triangle.

Hence it is proved that, $\Delta ABC$is an isosceles triangle.