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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 a triangle ABC with line segments cutting the internal angles at A and C in half (the angle bisectors). Extend these lines until they meet at a point I and draw perpendiculars from I to the three sides of ABC. (a) Using trigonometry or otherwise, show that these three perpendiculars all have the same length (call it r, the in radius). (b) Hence show that the line segment IB is also an angle bisector and that these all meet at I (the in centre). (c) A circle centred at I and with radius r (the in circle) 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).

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Also, what school does your son go to where they learn about Gergonne points? Wish I had gone there... – Potato Mar 4 '13 at 7:17

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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Thank you for your help @ Sarkar – Lucy Mar 2 '13 at 20:08

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