Let $ABCD$ be a parallelogram. An interior circle of the $ABCD$ is tangent to the lines $AB$ and $AD$ and intersects the diagonal $BD$ at $E$ and $F$. Prove that exists a circle that passes through $E$ and $F$ and is tangent to the lines $CB$ and $CD$.
Problem
Source: 2018 Argentina OMA L3 p3
Tags: geometry, circle, tangent, parallelogram
Neo-Pythagorean
20.06.2020 12:36
Let the interior circle $\omega$ touch $AB, AD$ at $M, N$ respectively. Name $\Omega$ the circle passing through $E$ and $F$ tangent to $CB$ at $P$. It is sufficient to prove that $\Omega$ is tangent to $CD$. Let the tangent from D to $\Omega$ touch $\Omega$ at $S$.Also, let $SD$ intersect $CB$ at $C'$. We will prove that $C'$$\equiv{C}$. We have: $$C'S=C'D+DS=C'D+\sqrt{DE*DF}=C'D+DN$$$$C'P=C'B+BP=C'B+\sqrt{BF*BE}=C'B+BM$$But $C'S=C'P$ (as tangents), so $C'D+DN=C'B+BM(1)$. Also: $$CD+DN=AB+(AD-AN)=(AM+MB)+(AD-AM)=BM+AD=BC+BM \Longleftrightarrow CD+DN=CB+BM(2)$$(1) and (2) give that $C'D-CD=C'B-CB$. Now, assume on the contrary that $C'$ belongs inside the segment $BC$: $$C'D-CD=C'B-CB \Longleftrightarrow C'D-CD=C'B-(C'C+C'B) \Longleftrightarrow CD= C'D+C'C$$This relationship violates the triangle inequality, yielding a contradiction. The other case (that C' lies outside the segment BC) is handled similarly. Thus, $C'$$\equiv{C}$ and we are done.
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