Question

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer 1

Let $BQ$ and $CP$ be the bisectors of $\angle ABC$ and $\angle ACB$ respectively, intersecting in the interior of $△ABC$ at $R$.

Let $BQ$ intersect side $AC$ in $Q$ and $CP$ intersect side $AB$ in $P$.

$\therefore$ By angle bisector theorem,

Since, $R$ lies on $BQ$, point $R$ is equidistant from $AB$ and $BC.$

Similarly, $R$ lies on $CP$ and is equidistant from $AC$ and $BC$.

So, $O$ is equidistant from $BC$ and $AC.$

Therefore, point $O$ is equidistant from all three sides $AB,BC$ and $CA$ of $△ABC$.