Question

Given: $\overline{\mathrm{AB}}\cong \overline{\mathrm{AC}}$, $\overline{\mathrm{AL}}$and$\overline{\mathrm{AM}}$ trisect $\angle \mathbf{BAC}$. (This means $\angle \mathbf{1}\cong \angle \mathbf{2}\cong \angle \mathbf{3}$)

Prove:$\overline{\mathrm{AL}}\cong \overline{\mathrm{AM}}$

Short answer

It is proved that, $\overline{AL}\cong \overline{AM}$.

Step-by-step solution

Step 1. Isosceles triangle theorem.

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

It is given that, $\overline{AB}\cong \overline{AC}$. Therefore, by isosceles triangle theorem, $\angle B\cong \angle C$.

Step 2. 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, $\angle B\cong \angle C$,$\overline{AB}\cong \overline{AC}$ and $\angle 1\cong \angle 3$, then by ASA postulate $\Delta BLA\cong \Delta CMA$.

Step 3. Description of step.

By congruent parts of congruent triangles, $\overline{AL}\cong \overline{AM}$.

Hence it is proved that, $\overline{AL}\cong \overline{AM}$.