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.
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 $R=R'$