Question

In the given figure, $\Delta$ABC and $\Delta$PBC are two isosceles triangles on the same base BC and vertices A and P are on the same side of BC. A and P are joined, then.

• A. $\angle BPA=\frac{1}{2}\angle BAC$
• B. $\angle BAP=\frac{1}{2}\angle BAC$
• C. $\angle CPA=\frac{1}{2}\angle BAC$
• D. $\angle BAP=2\angle BAC$

Answer 1

Answer: (B)

$\angle BAP=\frac{1}{2}\angle BAC$

In$△APB$ and $△APC$:

$AP=AC$

$BP=BC$

and have the same side $AP$

$\therefore △APC\equiv △APB$

We can say, $\angle BAP=\angle CAP$

and from the figure, $\angle BAC=\angle BAP+\angle CAP=2\angle BAP$

$\therefore \angle BAP=\frac{1}{2}\angle BAC$