Chapter 7.5, Problem 19WE

To determine

To prove: The triangles are similar for the given conditions.

Explanation of Solution

Given information:

  $\text{Δ}ABC$ is isosceles, it implies that,

  $\angle A=\angle C$

Consider the following figures two isosceles triangles.

  

From the above figure, it is given that $\text{Δ}ABC$ and $△DEF$ are isosceles triangle such that,

  $\angle B=\angle E$

It is required to prove that $\text{Δ}ABC-\text{Δ}DEF$

Since $\text{Δ}ABC$ is isosceles, it implies that,

  $\angle A=\angle C$

The sum of three angles of a triangle is 180°

This implies that,

  $\begin{array}{cc}\angle A+\angle B+\angle C& ={180}^{\circ }\\ \angle A+\angle B+\angle A& ={180}^{\circ }\\ 2\angle A& ={180}^{\circ }-\angle B\\ \angle A& =\frac{{180}^{\circ }-\angle B}{2}\end{array}$

This further implies that $\angle A\text{and}\angle C\text{both equals}\frac{{180}^{\circ }-\angle B}{2}$ .

In a similar way, since ?DEF is isosceles, it can be proved that $\angle D$ and $\angle F$ both equals $\frac{{180}^{\circ }-\angle E}{2}$ .

Since $\angle B=\angle E,$ , it follows that $\angle A=\angle D,\angle C=\angle F$ .

Thus using AAA similarity theorem, the triangles $\text{Δ}ABC$ and $△DEF$ are similar.