Problem 15: The converse of the butterfly theorem works here

by henderson, Jun 1, 2016, 11:58 AM

$$\bf\color{red}Problem \ 15 \ $$The circle $\omega$ passing through the vertices $B$ and $C$ of a triangle $ABC$ with $AB\ne AC$ intersects the sides $AB$ and $AC$ at the points $R$ and $S,$ respectively. Let $M$ be the midpoint of $BC.$ The line perpendicular to $MA$ at $A$ intersects $BS$ and $CR$ at $K$ and $T,$ respectively. If $AT=AK,$ prove that $MS=MR.$
$$\bf\color{red}My \ solution \ $$Applying the converse of the butterfly theorem to the quadrilateral $BSRC,$ we conclude that $OA\perp TK,$ where $O$ is the circumcenter of the quadrilateral $BRSC.$ Since $MA\perp TK,$ it follows that $O\in AM.$ If $O\not= M,$ then $AM\perp BC$ holds $($ because of $O\in AM$ and $OM\perp BC$ $)$ $,$ so $AB=AC,$ which is impossible. Therefore, $M$ and $O$ coincide. This leads to $MS=MR.$ $\quad$ $\square$
This post has been edited 7 times. Last edited by henderson, Mar 27, 2017, 2:45 PM

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I've just added the solution.

by henderson, Mar 27, 2017, 2:47 PM

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However, I have some suspicions that the solution can be wrong. Because this solution is very short and easy. So, if you see any mistake in the solution, please let me know.

by henderson, Mar 27, 2017, 2:54 PM

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

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