TITLE: Proving the existence of a triangle's incenter and Gergonne point
QUESTION [1 upvotes]: 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).

REPLY [1 votes]: 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.
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