Question

In the given figure, $ABC$ is an equilateral triangle and $P$ is any point in $AC$; then $BP>PA$?
State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

Since the triangle ABC is an equilateral triangle, each angle $={60}^o$

Clearly, from the figure, $\angle BAP>\angle ABP$ since $\angle BAP={60}^o$ and $\angle ABP$ is a part of ${60}^o$

This means, the sides opposite to the angles will also have the same relation.
$\therefore BP>PA$