In a triangle,locate a point in its interior which is equidistant from all the sides of the triangle.
(N/A) Consider a $\Delta ABC$.
Draw the angle bisector $'l'$ of $\angle B$.
Draw the angle bisector $'m'$ of $\angle C$.
Let the two angle bisectors $'l'$ and $'m'$ intersect at point $O$.
Since any point on the angle bisector of an angle is equidistant from the arms of the angle,point $O$ lies on the bisector of $\angle B$ (so it is equidistant from $AB$ and $BC$) and also on the bisector of $\angle C$ (so it is equidistant from $BC$ and $AC$).
Therefore,point $O$ is equidistant from all three sides ($AB$,$BC$,and $AC$) of $\Delta ABC$. This point $O$ is known as the incenter of the triangle.
Draw the angle bisector $'l'$ of $\angle B$.
Draw the angle bisector $'m'$ of $\angle C$.
Let the two angle bisectors $'l'$ and $'m'$ intersect at point $O$.
Since any point on the angle bisector of an angle is equidistant from the arms of the angle,point $O$ lies on the bisector of $\angle B$ (so it is equidistant from $AB$ and $BC$) and also on the bisector of $\angle C$ (so it is equidistant from $BC$ and $AC$).
Therefore,point $O$ is equidistant from all three sides ($AB$,$BC$,and $AC$) of $\Delta ABC$. This point $O$ is known as the incenter of the triangle.
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