# Prove that $PR$ tangents to the incircle of $\triangle{ABC}$

Incircle $$(I)$$ of $$\triangle{ABC}$$ tangents to $$AB$$ and $$AC$$ at $$M$$ and $$N$$ respectively. Let $$P$$ be any point lie on $$BC$$. $$AP$$ cuts $$CM$$ at $$Q$$. $$NQ$$ cuts $$AB$$ at $$R$$. Prove that $$PR$$ tangents to the incircle $$(I)$$. I got this problem from AoPS

There are some solution too but could someone give me the solution using normal elementary geometry? Or perhaps using Newton's theorem? I tried using this theorem and put the midpoint of $$AP$$ and $$CQ$$ by $$E$$ and $$F$$ respectively. But I don't know how I could prove that $$E-I-F$$ are collinear. Where should I apply Menelaus's theorem? Please help

If you don't want to use Pascal's or Brianchon's theorem, we can work around with simpler but algebraically heavier method.

I will prove an equivalent result, that in a tangent quadrilateral $$ABCD$$ lines $$AC$$, $$BN$$ and $$DM$$ intersect at one point.

I will assume known Menelaus's theorem, and formulae for inradius via tangent lengths for triangle and for tangent quadrilateral: $$r^2 = \frac{xyz}{x+y+z}=\frac{abc+abd+acd+bcd}{a+b+c+d}.$$

Our goal to find the ratio $$AQ/QC$$. First, let's find the length $$e=EM=EK$$. Since circle is inscribed both to quadrilateral $$ABCD$$ and triangle $$AED$$: $$r^2 = \frac{abc+abd+acd+bcd}{a+b+c+d} = \frac{aed}{a+e+d},\\ e = \frac{a b c+a b d+a c d+b c d}{a d-b c}.$$

Now consider the line $$BN$$ crossing the triangle $$AED$$: $$\frac{AB}{BE}\cdot\frac{EG}{GD}\cdot\frac{DN}{NA} = \frac{a+b}{e-b}\cdot\frac{e+d+x}{x}\cdot\frac da=1,\\ GD=x=\frac{a d (b+d) (c+d)}{c d (a+b)+a b c-a d^2}.$$

Consider the line $$BN$$ crossing the triangle $$ACD$$: $$\frac{AQ}{QC}\cdot\frac{CG}{GD}\cdot\frac{DN}{NA} = \frac{AQ}{QC}\cdot\frac{c+d+x}{x}\cdot\frac{d}{a}=1,\\ \frac{AQ}{QC} = \frac{a^2 (b+d)}{a b c+a b d+a c d+b c d}$$

This formula is symmetric relative to swapping $$b\leftrightarrow d$$, which means that segment $$DM$$ intersects $$AC$$ at the same point.

Finally, for your problem, we can choose point $$R'$$ on $$AB$$, so $$AR'PC$$ is a tangent circle, then since both $$NR$$ and $$NR'$$ pass through $$Q$$, we show that $$R=R'$$

• In the photo $AC, BN, MN$ is not concurrent did you mean other line? – user635988 Apr 30 at 7:04
• Ohh I see you mean $DM$. BTW could you explain about $AQ/QC$ why our goal is to find it? And in the end we found $AQ/QC$ but why it is imply that $DM$ intersects $AC$? – user635988 Apr 30 at 7:10
• If $BN$ divides $AC$ in ratio $k$ and $DM$ divides $AC$ in the same ratio, then 3 segments are concurrent. – Vasily Mitch Apr 30 at 9:18
• After we did all of the math to find where $BN$ intersects $AC$, we can repeat it again with $DM$. All our arguments will be the same, except when we need to write $b$, we will write $d$ and vice versa. As a result, we will end up with a similar formula where $b$ and $d$ are swapped. However, since $b$ and $d$ enter the formula for $AQ/QC$ in a symmetric manner, the answer will be exactly the same. – Vasily Mitch Apr 30 at 9:20
• Ohhh so if I'm using $DM$ then let $x$ be the intersecting point of $DM$ and $AC$ and apply Menelaus's to $∆AEC$ and find $AX/XC$ is that right? – user635988 Apr 30 at 11:37