Chapter 2.6, Problem 25WE

To determine

To prove: $\angle DBC\cong \angle FEB$

Explanation of Solution

Given:

  $m\angle DBA={45}^o$ and $m\angle DEB={45}^o$

From the figure,

  $m\angle DBA+m\angle DBC={180}^o$ and

  $m\angle DEB+m\angle FEB={180}^o$

Also,

  $m\angle DBC={180}^o-{45}^0={135}^o$ and

  $m\angle DBC={180}^o-{45}^0={135}^o$

So, from above condition

  $m\angle DBC=m\angle FEB$

It means

  $m\angle DBC\cong m\angle FEB$ (since equal are congruent angles)

Hence, it is proved that

  $m\angle DBC\cong m\angle FEB$