Chapter 1.4, Problem 11PSB

To determine

To Show: $\angle DBC\cong \angle XFE$

Explanation of Solution

Given information :

  $\begin{array}{l}\angle ABD=10°\\ \angle ABC=100°\\ \angle EFY=70°20^{\prime }\\ \angle XFY=19°40^{\prime }\end{array}$

  

Proof:

Here, we have

  $\begin{array}{l}\angle ABD=10°\\ \angle ABC=100°\\ \angle EFY=70°20^{\prime }\\ \angle XFY=19°40^{\prime }\\ \because \angle DBC=\angle ABC-\angle ABD\\ \Rightarrow \angle DBC=100°-10°\\ \Rightarrow \angle DBC=90°\left(1\right)\\ \text{Also}\\ \because \angle XFE=\angle XFY+\angle EFY\\ \Rightarrow \angle XFE=19°40^{\prime }+70°20^{\prime }\\ \Rightarrow \angle XFE=\left(19°+70°\right)+\left(40^{\prime }+20^{\prime }\right)\\ \Rightarrow \angle XFE=89°+60^{\prime }\\ \Rightarrow \angle XFE=89°+1°\\ \Rightarrow \angle XFE=90°\left(2\right)\\ \text{From (1) and (2), we get}\\ \angle DBC\cong \angle XFE\end{array}$