Question

In the figure $\Delta APQ,P-B-C-Q$ and $AB=AC=PB=CQ$. Find the angle congruent to $\angle PAQ$.

• A. $\angle$ACP
• B. $\angle$ABP
• C. $\angle$APC
• D. $\angle$BAQ

Answer 1

Answer: (B)

$\angle$ABP

In $△ABC$, $AB=AC$

$⟹\angle ABC=\angle ACB=x\left(\text{say}\right)$ ......$\left(i\right)$ $\left[\text{Angles opposite to equal sides are themselves equal}\right]$

Similarly, in $△ABP$ and $△ACQ$, $AB=PB\text{and}AC=CQ$

$⟹\angle APB=\angle BAP=y\left(\text{say}\right)$ and $\angle AQC=\angle QAC=z\left(\text{say}\right)$ ...... $\left(ii\right)$

Now, in $△ABC,$

$\angle ABC+\angle ACB+\angle BAC={180}^o$

$⟹\angle BAC={180}^o-2x$ ........... $\left(iii\right)$

In $△ABP,$

$\angle ABP+\angle APB+\angle BAP={180}^o$

$⟹\angle ABP={180}^o-2y$ .......... From $\left(ii\right)$

Also, $\angle ABP={180}^o-x$ $[\because \angle ABC+\angle ABP={180}^o]$ ...... $\left(iv\right)$

$⟹180-2y={180}^o-x$

$⟹y=\frac{x}{2}$

$⟹\angle APB=\angle BAP=\frac{x}{2}=\angle AQC=\angle QAC$ ........... From $\left(ii\right)$

Now, $\angle PAQ=\angle PAB+\angle BAC+\angle CAQ$

$=\frac{x}{2}+{180}^o-2x+\frac{x}{2}$

$={180}^o-x=\angle ABP$ ......... From $\left(iv\right)$

Hence, $\angle PAQ=\angle ABP$