Question

In the following figure, $ABC$ is an equilateral triangle and $P$ is any point in $AC$; then Is $BP>PC$ ?

Say true or false.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient
  

Answer 1

The correct option is A True

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

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

This means, the sides opposite to the angles will also have the same relation.

$\therefore BP>PC$