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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

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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.

enter image description here 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 $R=R'$

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  • $\begingroup$ In the photo $AC, BN, MN$ is not concurrent did you mean other line? $\endgroup$ – user635988 Apr 30 at 7:04
  • $\begingroup$ 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 $AC$? $\endgroup$ – user635988 Apr 30 at 7:10
  • $\begingroup$ If $BN$ divides $AC$ in ratio $k$ and $DM$ divides $AC$ in the same ratio, then 3 segments are concurrent. $\endgroup$ – Vasily Mitch Apr 30 at 9:18
  • $\begingroup$ 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. $\endgroup$ – Vasily Mitch Apr 30 at 9:20
  • $\begingroup$ 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? $\endgroup$ – user635988 Apr 30 at 11:37

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