Chapter 7.2, Problem 11PSB

To determine

To prove : The altitude to base of an isosceles triangle is also a median to the base.

Explanation of Solution

Given information : The following information has been given

ABC is an isosceles triangle with AD as altitude

  $\angle B=\angle C$

  $\angle ADB=\angle ADC={90}^0$

  $\overrightarrow{AB}=\overrightarrow{AC}$

Formula used : If any two angles of a triangle are congruent, and the corresponding sides are also congruent, then the triangle can be proved congruent by ASA by adding the angles to ${180}^0$ and proving it to be similar first

Proof : We know that in triangles ABD and ACD,

  $\angle B=\angle C$

  $\angle ADB=\angle ADC={90}^0$

  $\overrightarrow{AB}=\overrightarrow{AC}$

Thus, we can say that by ASA, $\Delta ABD\cong \Delta ACD$

Now, by virtue of congruency, we can say that

  $\overrightarrow{BD}=\overrightarrow{CD}$

Hence, proved that the altitude to base of an isosceles triangle is also a median to the base