Question

In fig and fid seg AB $\cong$ AC, seg BP $\cong$ PC, $\angle A={50}^{\circ },\angle P={110}^{\circ }.$ Then find $\angle ABP$ and $\angle ACP$ and hence find the relation between them

• A. $\angle ABP={100}^{\circ }$ or ${30}^{\circ }$
• B. $\angle ABP={30}^{\circ }$
• C. $\angle ABP={100}^{\circ }$
• D. None of these

Answer 1

The correct option is A $\angle ABP={100}^{\circ }$ or ${30}^{\circ }$

Given: $AB=AC$ and $BP=PC$
Join AP
Now, In $△ABP$ and $△APC$
$AB=AC$ (Given)
$BP=PC$ (Given)
$AP=AP$ (Common)
Thus, $△ABP\cong △ACP$
Hence, $\angle BAP=\angle CAP=\frac{1}{2}\angle A={25}^{\circ }$
$\angle ABP=\angle ACP$
In Case I,
$\angle ABP=\frac{1}{2}\angle P={55}^{\circ }$
In Case II,
$\angle ABP=\frac{1}{2}(ext.\angle P)={125}^{\circ }$
Now, In $△APB$
Sum of angles = 180
$\angle ABP+\angle APB+\angle PAB=180$
In case I,
$\angle ABP+25+55=180$
$\angle ABP={100}^{\circ }$
In case II,
$\angle ABP+25+125=180$
$\angle ABP={30}^{\circ }$
therefore, $\angle ABP={100}^{\circ }or{30}^{\circ }$