efoski1687 wrote: A circle is tangent to sides $AD,\text{ }DC,\text{ }CB$ of a convex quadrilateral $ABCD$ at $\text{K},\text{ L},\text{ M}$ respectively. A line $l$, passing through $L$ and parallel to $AD$, meets $KM$ at $N$ and $KC$ at $P$. Prove that $PL=PN$. let $Q$ be the intersection of the extensions of the lines $AD$ and $CB$ and $P$ be the intersection of the extensions of the lines $KL$ and $BC$. Obviously we have: $(QCMP)=-1$ so $K(QCMP)=-1$ and therfore: $K(\infty PNL)=-1$ which implies that:$PN=PL$

Let's draw a line that pass from $C$ and parallel to $AD$ . Inersection of this line and $KM$ is $J$. Equal tangent lengts $DK=DL=x$ and $CL=CM=y$. Since triangle $CJM$ is isosceles, $CJ=y$. From similarity of $CLP \sim CDK$, $PL = \frac{xy}{x+y}$ ... (1) and from similarity of $KPN \sim KCJ$, $PN = \frac{xy}{x+y}$ ... (2) By (1) and (2), clearly $PL = PN$.

One can easily see that $KC$ is the $C$ - symmedian in $\triangle KLM$. Let $X$ be the intersection of $KC$ with $LN$. We have $\frac{LX}{XM}=\frac{KL^2}{KM^2}$ Also, $\angle KML=\angle KLD=\angle LKD=\angle KLN$ So we have $\triangle KLN \sim \triangle KML$ Using this similarity we can easily get $KM\cdot KN=KL^2$ Now we apply Menelaus Theorem to $\triangle MNL$ $1 =\frac{KM}{KN}\frac{PN}{PL}\frac{LX}{XM}= \frac{PN}{PL}\frac{KL^2}{KM^2}\frac{KM}{KN}= \frac{PN}{PL}\frac{KL^2}{KM\cdot KN} =\frac{PN}{PL}$ $PN=PL$