Chapter 3.8, Problem 4PSA

To determine

The congruency of $\angle CDE\cong \angle ABF$ is to be proved

Explanation of Solution

Given information : The following information has been given

  $\overrightarrow{AE}\cong \overrightarrow{CF}$

  $\overrightarrow{AB}\cong \overrightarrow{CD}$

  $\angle BFA\cong \angle DEC={90}^0$

Formula used : Any two triangles are said to be congruent by SSS (side-side-side) property if they have all 3 sides congruent

Proof : We know that $\overrightarrow{AE}\cong \overrightarrow{CF}$ , hence, we add EF on both sides. Thus, we get

  $\overrightarrow{AF}\cong \overrightarrow{CE}$

Thus, we use Pythagoras theorem on the triangles ABF and CDE. We know that

  $\overrightarrow{AB}\cong \overrightarrow{CD}$ , $\overrightarrow{AF}\cong \overrightarrow{CE}$ and $\angle BFA\cong \angle DEC={90}^0$

Thus, we now get

  $\overrightarrow{FB}\cong \overrightarrow{ED}$

Hence, now we have all 3 sides congruent. Thus, we can say that

  $\Delta CDE\cong \Delta ABF$ by SSS property

Now, we know that congruent triangles have congruent angles. Thus, we can also say that

  $\angle CDE\cong \angle ABF$

Hence, required congruency is proved