Problem 63: Concurrent lines

by henderson, Mar 26, 2017, 12:23 PM

$\color{red}\bf{Problem \ 63}$ Let $O$ be the circumcenter of an acute $\triangle ABC.$ The points $E$ and $F$ are chosen on the lines $AB$ and $AC,$ respectively, such that $O$ is the midpoint of $EF.$ Let $D$ be the intersection point of $AO$ $($ other than $A$ $)$ with the circumcircle of $\triangle ABC.$ Prove that the lines $EF, BC$ and the tangent to the circumcircle of $\triangle ABC$ at $D$ are concurrent.
$\color{red}\bf{My \ solution}$ Denote by $l$ the tangent to the circumcircle of $\triangle ABC$ at $D$ and let $\{ S \}=BC \cap l.$ Applying the butterfly theorem to the degenerated quadrilateral $BDCD,$ we obtain that, if $m$ is the line passing through the point $S$ such that $m \perp OS,$ then the intersection points of $BD$ and $CD$ with the line $m$ have equal distances from the point $S.$ In other words, if $\{ E_2 \}=BD\cap m$ and $\{ F_2 \}=CD\cap m,$ then $SE_2=SF_2.$

$\color{blue}\textbf{Claim.}$ $\quad$ $\triangle DE_2F_2 \sim \triangle AEF.$

$\color{blue}\textbf{Proof.}$ $\quad$ Since $l$ is tangent to the circumcircle of $\triangle ABC,$ $\angle EAO=\angle E_2DS$ and $\angle FAO=\angle F_2DS$ hold. On the other hand, sines' law imply
$\frac{E_2D}{\sin \angle E_2SD}=\frac{E_2S}{\sin \angle E_2DS} \ \text{and} \ \frac{F_2D}{\sin \angle F_2SD}=\frac{F_2S}{\sin \angle F_2DS}$ $\Longrightarrow$

$\frac{E_2D}{F_2D}=\frac{\sin \angle F_2DS}{\sin \angle E_2DS}=\frac{\sin \angle FAO}{\sin \angle EAO} .$ Moreover, $\frac{EA}{FA}=\frac{\sin \angle FAO}{\sin \angle EAO}$ in $\triangle AEF.$
Thus, we obtain $\frac{E_2D}{F_2D}=\frac{EA}{FA}.$ Taking into account the fact that
$\angle E_2DF_2=\angle E_2DS+\angle F_2DS= \angle EAO+\angle FAO=\angle EAF,$ we conclude
$\triangle DE_2F_2 \sim \triangle AEF ,$ as desired. $\quad$ $\square$

Let $\{ E_1 \}=OS\cap AB.$ Since $\angle E_2SE_1=\angle E_2BE_1,$ the quadrilateral $E_2SBE_1$ is cyclic $\Longrightarrow$ $\angle SE_2B=\angle SE_1B.$ As $\angle SE_2B=\angle F_2E_2D=\angle FEA=\angle OEA,$ we obtain $\angle SE_1B=\angle OEA,$ which means that $E$ and $E_1$ coincide.
Therefore, the lines $l, BC$ and $EO$ are concurrent at the point $S$ $\Longrightarrow$ $l, BC$ and $EF$ are concurrent at the point $S.$ $\quad$ $\square$

This post has been edited 4 times. Last edited by henderson, Mar 30, 2017, 4:49 PM