Chapter 7.2, Problem 12PSB

To determine

To prove : The segments drawn from the midpoint of the base of an isosceles triangle and perpendicular to the legs are congruent if they terminate at the legs.

Explanation of Solution

Given information : The following information has been given

Let’s consider the triangle ABC with midpoint of base at D

Perpendicular Segments DE on AB and DF on AC are drawn

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 DEB and DFC

  $\angle B=\angle C$

  $\angle DEB=\angle DFC={90}^0$

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

Thus, we can prove that $\Delta DEB\cong \Delta DFC$ by ASA method

Hence, by virtue of congruency, $\overrightarrow{DE}\cong \overrightarrow{DF}$

Thus, proved that the segments drawn from the midpoint of the base of an isosceles triangle and perpendicular to the legs are congruent if they terminate at the legs