Question

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

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

Short answer

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

Step-by-step solution

Step 1. Description of step.

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

Step 2. 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, $\overline{AL}\cong \overline{AM}$.

Step 3. 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 $\Delta ABL\cong \Delta ACM$.

Step 4. Description of step.

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

Step 5. Description of step.

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

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