# Proving that given a triangle $ABC$ the line $AB$ is parallel to $CK$

Supposing a triangle $$ABC$$, where $$|AC|\neq |BC|$$, denote the incentre $$I$$ and the points of tangency between the inscribed circle and $$BC$$,$$CA$$,$$AB$$ to be $$D,E,F$$ respectively. $$M$$ is the midpoint of $$AB$$. $$K$$ is the intersection of the perpendicular to $$CM$$ passing through $$I$$, and line $$DE$$. The task is to prove that $$CK||AB$$.

I have realized that $$DE$$ is a side of the Gergonne triangle of $$ABC$$, but I haven't been able to connect it with angle bisectors or anything pertaining to the incentre, nor with medians. I have also tried constructing the circumscribed circle and looking at the perpendicular bisectors but without much success.

It's well known that $$DE, IF, CM$$ are concurrent. Call this point $$X$$.

Then $$CX\perp IK$$, $$CI\perp XK$$, meaning $$I$$ is the orthocenter of $$CXK$$ and the result follows.

Another interesting thing is that $$\triangle CXK$$ is self-polar wrt the incircle.

Edit: The following proof that $$DE,IF,CM$$ are concurrent is not mine. It's taken from Evan Chen's EGMO.

Let $$X$$ be the intersection of $$DE$$ and $$FI$$ and let $$A',B'$$ be points on $$CA$$ and $$CB$$ such that $$A'B'\parallel AB$$ and $$A'B'$$ passes through $$X$$.

Now $$E$$ is the foot of $$I$$ onto $$CA'$$, $$X$$ is the foot of $$I$$ onto $$A'B'$$ ($$B'A'\parallel BA$$ and $$IF\perp AB$$, so $$IF\perp AB$$), and $$D$$ is the foot of $$I$$ onto $$CB'$$. Then $$EXD$$ is in fact the Simson line of $$I$$ wrt $$\triangle CA'B'$$, which means that $$I$$ is on the circumcircle of $$\triangle CA'B'$$.

Since $$CI$$ is an angle bisector of $$\angle A'CB'$$, $$I$$ is the arc midpoint of arc $$B'A'$$, and so $$XI$$ is the perpendicular bisector of $$B'A'$$. This means $$X$$ is the midpoint of $$B'A'$$, and so taking a homothety at $$A$$ which takes $$A'B'\mapsto AB$$ shows that the $$C$$-median passes through $$X$$ as well.

Once seeing $$I$$ is the orthocenter of $$\triangle CXK$$, $$CK\perp IX$$ but we already know $$IX\perp AB$$ so $$CK\parallel AB$$.

• Thanks! Would you care to elaborate on why $DE,IF,CM$ are concurrent? How exactly does the result follow from $I$ being the orthocenter of $CXK$? Mar 16 at 13:00
• You skip a lot of steps there, it would be nice if you expanded on your reasoning about the concurrency and how exactly does the result OP has asked for follow from your final point. Mar 16 at 18:35
• I have expanded my post. Mar 16 at 19:01