# Proving the existence of a triangle's incenter and Gergonne point

I have no idea on how to solve this question so can someone please assist me. My son brought it from school and he is really struggling with the question.

Consider $$\triangle ABC$$. Extend the internal angle bisectors at $$A$$ and $$C$$ until they meet at a point $$I$$, and draw perpendiculars from $$I$$ to the three sides of $$\triangle ABC$$.

a. Using trigonometry or otherwise, show that these three perpendiculars all have the same length (call it $$r$$, the inradius).

b. Hence show that $$\overline{IB}$$ is also an angle bisector, and that these all meet at $$I$$ (the incentre).

c. A circle centered at $$I$$ and with radius $$r$$ (the incircle) is tangent to each side of the triangle. Show that the lines joining the points of contact to the opposite vertices all meet at a single point (called the Gergonne point).

This is the best diagram I found on the internet in image search (next time, try to provide a diagram with the question). It calls the point inside $$O$$ instead of $$I$$. Let the point on $$AC, CB, BA$$ be $$X, Y, Z$$ respectively. Then all you have to show for the first part is that triangle $$AOX$$ is congruent to triangle $$AOZ$$. This is easy because they are both right angled, have equal half angles at $$A$$ and share a common side $$AO$$. So $$OX = OZ$$. You can prove that $$OZ = OY$$ with a similar argument and so all perpendiculars have the same length. For the next part, you can show that triangle $$BOZ$$ is congruent to $$BOY$$. So the two angles at $$B$$ are equal and hence $$OB$$ is an angle bisector. A tangent to a circle at point $$P$$ is perpendicular to $$OP$$ and passes through $$P$$. All the sides satisfies this, so they are tangent to the incircle. I am not sure how to do the Gergonne point question.