Chapter 12.1, Problem 42E

To determine

To prove: the exterior angle theorem and sum of angle $A$ and $B$ gives angle $C$ .

Answer to Problem 42E

  $m\angle A+m\angle B=m\angle BCD$ .

Explanation of Solution

Given:

  

Concept used:

Sum of the three interior angles of triangle is always ${180}^0$ .

Sum of two nonadjacent angle is equal to the exterior angle.

The sumof two adjacent angle made by two intersection line is linear pair and there sum is ${180}^0$ .

Calculation:

According to the given diagram:

  $ABC$ is a triangle and $\angle BCD$ is an Exterior angle shows in diagram.

From the given diagram it is also cleared that the $m\angle BCA\text{ and }m\angle BCD$ form a linear pair.

  $m\angle BCA\text{ +}m\angle BCD={180}^0.$ ............ $\left(i\right)$ .

Sum of the three interiors angle of the triangle is also $180$ .

  $m\angle A+m\angle B+m\angle BCA={180}^0.$ ............ $\left(ii\right)$ .

Equating equation $\left(i\right)$ and $\left(ii\right)$ :

  $\begin{array}{l}m\angle BCA\text{ +}m\angle BCD=m\angle A+m\angle B+m\angle BCA.\\ \Rightarrow m\angle A+m\angle B=m\angle BCD.\end{array}$

Hence, $m\angle A+m\angle B=m\angle BCD$ .