1. **Problem Statement:** (i) Show that any two angle bisectors of a triangle meet in a point. (ii) Show that the three angle bisectors of a triangle meet in a point. 2. **Key Concept:** The angle bisector of a triangle is a line that divides an angle into two equal parts. 3. **Step 1: Consider a triangle $\triangle ABC$ with angle bisectors from vertices $A$ and $B$** - Let the angle bisector from $A$ meet side $BC$ at $D$. - Let the angle bisector from $B$ meet side $AC$ at $E$. 4. **Step 2: Show that these two bisectors intersect at a point $I$ inside the triangle** - Since $AD$ and $BE$ are lines inside the triangle, they must intersect at some point $I$ (by the property of lines in a plane). 5. **Step 3: Show that $I$ lies on the bisector of angle $C$ as well** - By the Angle Bisector Theorem, point $I$ is equidistant from the sides of the angles it bisects. - Since $I$ lies on bisectors of $A$ and $B$, it is equidistant from sides $AB$, $AC$, and $BC$. - Therefore, $I$ must also lie on the bisector of angle $C$. 6. **Step 4: Conclusion** - The three angle bisectors meet at a single point $I$, called the incenter of the triangle. - This point is equidistant from all sides of the triangle. **Summary:** - Any two angle bisectors intersect at a point inside the triangle. - This point lies on the third angle bisector as well. - Hence, all three angle bisectors meet at a single point, the incenter. This completes the proof.