Chapter 12, Problem 37CR

To determine

To calculate:Themeasure of angle DEC in an isosceles triangle ABC.

Answer to Problem 37CR

The measure of angle DECis${20}^{\circ }$ .

Explanation of Solution

Given information:

  $\overline{BC}$ is the base of isosceles triangle ABC.

  $\overline{DE}$ is the base of isosceles triangle AED.

  $\angle BAE={40}^{\circ }.$

Calculation:

  

The interior angles of triangle add up to ${180}^{\circ }$

  $\angle x+{40}^{\circ }$ is supplementary of $\angle AEB$ .

  $\angle y+\angle z$ is supplementary of $\angle AEB$ .

  $\begin{array}{l}\angle x+{40}^{\circ }=\angle y+\angle z\\ x+{40}^{\circ }-y=z\\ z=x+{40}^{\circ }-y\end{array}$

  $\begin{array}{l}\angle EDC={180}^{\circ }-y\\ x+\angle EDC+z={180}^{\circ }\\ x+{180}^{\circ }-y+z={180}^{\circ }\\ x-y+z=0\\ y-x=z----Equation1\\ z=x+40^{\circ }-y\\ y-x=x+40^{\circ }-y\\ 2y-2x={40}^{\circ }\\ y-x={20}^{\circ }----Equation2\end{array}$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}z=y-x={20}^{\circ }\\ \angle DEC={20}^{\circ }\end{array}$