Question

a. Given: $\overline{\mathrm{AB}}\cong \overline{\mathrm{AC}}$; $\overline{\mathrm{BD}}\perp \overline{\mathrm{AC}}$; $\overline{\mathrm{CE}}\perp \overline{\mathrm{AB}}$.

Prove:$\overline{\mathrm{BD}}\cong \overline{\mathrm{CE}}$

b. The result you proved in part (a) can be stated as a theorem about certain altitudes. State this theorem in your own words.

Short answer

a. It is being given that$\overline{\mathrm{AB}}\cong \overline{\mathrm{AC}}$,$\overline{\mathrm{BD}}\perp \overline{\mathrm{AC}}$ and $\overline{\mathrm{CE}}\perp \overline{\mathrm{AB}}$.

As, $\overline{\mathrm{BD}}\perp \overline{\mathrm{AC}}$, therefore $\angle \mathbf{BDA}\mathbf{=}\mathbf{90}\mathbf{°}$.

As, $\overline{\mathrm{CE}}\perp \overline{\mathrm{AB}}$, therefore $\angle \mathbf{CEA}\mathbf{=}\mathbf{90}\mathbf{°}$.

Therefore, it can be said that $\angle \mathbf{BDA}\cong \angle \mathbf{CEA}$.

In the triangles$\mathrm{\Delta }\mathbf{BAD}$ and $\mathrm{\Delta }\mathbf{CAE}$, it can be noticed that:

$\overline{\mathrm{AB}}\cong \overline{\mathrm{AC}}\left(\mathrm{given}\right)$

$\angle \mathbf{BDA}\cong \angle \mathbf{CEA}$

$\angle \mathbf{BAD}\cong \angle \mathbf{CAE}\mathbf{}\left(\mathrm{reflexive}\mathrm{property}\right)$

Therefore, the triangles$\mathrm{\Delta }\mathbf{BAD}$ and$\mathrm{\Delta }\mathbf{CAE}$ are the congruent triangles by AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $\overline{\mathrm{BD}}\cong \overline{\mathrm{CE}}$.

b. The theorem is that in isosceles triangle, the altitudes drawn from the base angles are congruent.

Step-by-step solution

Part a. Step 1. Observe the given diagram.

The given diagram is:

Part a. Step 2. Description of step.

It is being given that$\overline{AB}\cong \overline{AC}$,$\overline{BD}\perp \overline{AC}$ and $\overline{CE}\perp \overline{AB}$.

As, $\overline{BD}\perp \overline{AC}$, therefore $\angle BDA=90°$.

As, $\overline{CE}\perp \overline{AB}$, therefore $\angle CEA=90°$.

Therefore, it can be said that $\angle BDA\cong \angle CEA$.

Part a. Step 3. Description of step.

In the triangles$\mathrm{\Delta }BAD$ and width="47" height="19" role="math">ΔCAE, it can be noticed that:

$\overline{AB}\cong \overline{AC}\left(\text{given}\right)$

$\angle BDA\cong \angle CEA$

$\angle BAD\cong \angle CAE\left(\text{reflexiveproperty}\right)$

Therefore, the triangles$\mathrm{\Delta }BAD$ and$\mathrm{\Delta }CAE$ are the congruent triangles by AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $\overline{BD}\cong \overline{CE}$.

Part b. Step 1. Observe the given diagram.

The given diagram is:

Part b. Step 2. Description of step.

It is being given that$\overline{AB}\cong \overline{AC}$,$\overline{BD}\perp \overline{AC}$ and $\overline{CE}\perp \overline{AB}$.

As, $\overline{BD}\perp \overline{AC}$, therefore$BD$is an altitude drawn from the vertex $B$.

As, $\overline{CE}\perp \overline{AB}$, therefore$CE$is an altitude drawn from the vertex $C$.

As, $\overline{BD}\cong \overline{CE}$, thereforeit can be said that the altitudes drawn from the vertex$B$and$C$are congruent.

As, $\overline{AB}\cong \overline{AC}$, therefore the triangle $\mathrm{\Delta }ABC$is an isosceles triangle.

In the isosceles triangle $\mathrm{\Delta }ABC$, the angles $\angle B$and$\angle C$are the base angles.

As, $\overline{BD}\cong \overline{CE}$, therefore in words it can be stated that in isosceles triangle, the altitudes drawn from the base angles are congruent.

Part b. Step 3. Description of step.

The theorem is that in isosceles triangle, the altitudes drawn from the base angles are congruent.