Chapter 3.3, Problem 21PSC

To determine

To prove : The congruency of $\angle 1$ with respect to $\angle 2$ .

Explanation of Solution

Given information : In this question, we have given that $\overline{AE}\cong \overline{FC}$ , $\overline{FB}\cong \overline{DE}$ and $\angle CFB\cong \angle AED$ .

Formula used: SAS Congruency rule.

Proof: As, it is given that $\overline{AE}\cong \overline{FC}$ , Adding $\overline{EF}$ on both the sides, we get,

  $\begin{array}{l}\Rightarrow \overline{AE}+\overline{EF}\cong \overline{FC}+\overline{EF}\\ \Rightarrow \overline{AF}\cong \overline{CE}\end{array}$

Now, $\angle CFB+\angle BFA=180°$ and $\angle AED+\angle DEC=180°$

  $\Rightarrow \angle CFB=180°-\angle BFA$ and $\angle AED=180°-\angle DEC$

Now, as $\angle CFB\cong \angle AED$ (Given)

  $\Rightarrow \angle CFB=180°-\angle BFA$ and $\Rightarrow \angle CFB=180°-\angle DEC$

  $\Rightarrow 180°-\angle BFA=180°-\angle DEC$

  $\Rightarrow \angle BFA=\angle DEC$

Now, In $\Delta AFBand\Delta CED$

  $\overline{FB}\cong \overline{DE}$ (Proved given)

  $\angle BFA=\angle DEC$ (Proved above)

  $\overline{AF}\cong \overline{CE}$ (Proved above)

  $\therefore \Delta AFB\cong \Delta CED$ (BY SAS Congruency rule)

  $\Rightarrow \angle 1\cong \angle 2$ (BY CPCT)