Chapter 3.8, Problem 5PSA

To determine

To Prove: The altitude to the base of an isosceles triangle divides the triangle into two congruent triangles

Explanation of Solution

Given information : We are given an isosceles triangle and its altitude.

Suppose ABC is the given isosceles triangle with AB and AC as its legs and AD as its altitude as shown in figure.

  

Concept used : RHS (Right angle − Hypotenuse- Side) Congruency rule.

Proof :

As per the question and drawn sketch we have ABC as an isosceles triangle where $\overline{AB}\cong \overline{AC}$ and AD is the altitude. We know that altitude of a triangle measure $90°$

Considering $\Delta ADB\text{ and }\Delta ADC$ we have:

  $\begin{array}{l}\overline{AB}\cong \overline{AC}\mathrm{............................................}\text{(given)}\\ \\ \overline{AD}\cong \overline{AD}\mathrm{............................................}\text{(common)}\\ \\ \angle ADB\cong \angle ADC\text{ = 90}°\mathrm{............................}\text{(given as altitude)}\end{array}$

So, using RHS (Right angle − Hypotenuse- Side) Congruency rule we can say that:

  $\Delta ADB\text{ and }\Delta ADC$ are congruent

Or $\Delta ADB\cong \Delta ADC$

Which proves the required result.