Question

Given: $\overline{\mathrm{AB}}\cong \overline{\mathrm{AC}};\overline{\mathrm{BN}}\mathbf{\perp }\overline{\mathrm{AC}}\mathbf{;}\mathbf{}\overline{\mathrm{CM}}\perp \overline{\mathrm{AB}}$ .

Explain how you could prove that $\Delta \mathbf{ABN}\cong \Delta \mathbf{ACM}$.

Short answer

In the triangles $\Delta ABN$and $\Delta ACM$, it can be noticed that:

$$\begin{gathered} AB=AC\left(given\right) \\ \angle A=\angle A\left(reflexiveproperty\right) \\ \angle ANB=\angle AMC=90° \end{gathered}$$

Therefore, the triangles $\Delta ABN$and $\Delta ACM$are the congruent triangles by using$AAS$ postulate.

Therefore, $\Delta ABN\cong \Delta ACM$can be proved by $AAS$postulate.

Step-by-step solution

- Observe the given diagram.

The given diagram is:

- Description of step.

In the triangles $\Delta ABN$and $\Delta ACM$, it can be noticed that:

$$\begin{gathered} AB=AC\left(given\right) \\ \angle A=\angle A\left(reflexiveproperty\right) \\ \angle ANB=\angle AMC=90° \end{gathered}$$

Therefore, the triangles$\Delta ABN$ and$\Delta ACM$ are the congruent triangles by using$AAS$ postulate.

- Description of step.

Therefore, the triangles$\Delta ABN$ and$\Delta ACM$ are the congruent triangles by using$AAS$ postulate.

Therefore, $\Delta ABN\cong \Delta ACM$can be proved by $AAS$postulate.