Chapter 6, Problem 15RP

To determine

To prove : $\overleftrightarrow{BA}$ is perpendicular to $\overleftrightarrow{PQ}$

Explanation of Solution

Given information : Here, we have given that $\overline{PC}\cong \overline{QC}$ , A is the midpoint of $\overline{PQ}$ and $\angle PCB\cong \angle QCB$ .

Proof: As, A is the midpoint of $\overline{PQ}$ ,

  $\Rightarrow \overline{AP}\cong \overline{AQ}$

Now, In $\Delta PAC$ and $\Delta QAC$ ,

  $\overline{AP}\cong \overline{AQ}$ (Proved above)

  $\overline{PC}\cong \overline{QC}$ (Given)

  $\overline{AC}\cong \overline{AC}$ (Common)

  $\Rightarrow \Delta PAC\cong \Delta QAC$ (By SSS Congruency Rule)

  $\therefore \angle PAC\cong \angle QAC$ (By Corresponding Part of Congruent Triangles)

Now, $\angle PAC+\angle QAC=180°$ (Linear Pair Angles)

  $\begin{array}{l}\Rightarrow \angle PAC+\angle PAC=180°\\ \Rightarrow 2(\angle PAC)=180°\\ \Rightarrow \angle PAC=\frac{180°}{2}\\ \Rightarrow \angle PAC=90°\end{array}$

  $\therefore \overleftrightarrow{BA}\perp \overleftrightarrow{PQ}$