DX bisects segment EF, <XEA =<XFA = 90^o, circumcircle, altitude

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DX bisects segment EF, <XEA =<XFA = 90^o, circumcircle, altitude

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Source: 2014 Thailand October Camp Geometry Exam p1

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#1 h Oct 22, 2020, 9:00 PM

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Let $O$ be the circumcenter of an acute $\vartriangle ABC$ which has altitude $AD$ . Let $AO$ intersect the circumcircle of $\vartriangle BOC$ again at $X$ . If $E$ and $F$ are points on lines $AB$ and $AC$ such that $\angle XEA = \angle XFA = 90^o$ , then prove that the line $DX$ bisects the segment $EF$ .

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#2 h Oct 23, 2020, 12:15 PM • 1 Y

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Let $M=XE\cap BC , N=XF\cap BC$
$\angle{XCF}=\angle{CAO}+\angle{CXO}$
$=\angle{CAO}+\angle{CBO}$
$=\angle{C}$
$=\angle{NCF}$
So $F$ is the midpoint of $XN$ . similary , $E$ is the midpoint of $XM$
Therefore $A$ is the circumcenter of $\Delta XMN$ so $D$ is the midpoint of $MN$ and by taking homothity we are done.
Remark: I think it could be done by complex bash but I couldn't compute $x$ :wallbash_red:

:wallbash_red:

@below Nice idea

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#3 h Oct 23, 2020, 6:25 PM • 1 Y

Y by Muaaz.SY

For complex bash you could just say that x=ta and t is real and then compute all other points is simple and assume x,d and the midpoint of EF are collinear and this way you find t. Now only need to show that XOBC lie on a common circle which is rather simple

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#4 h Mar 28, 2022, 6:58 AM

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Note that $\triangle AEF\sim\triangle ABC$ and by law of sine we have

$\frac{CF}{CA}=\frac{CF}{CX}\cdot\frac{CX}{OC}\cdot\frac{OC}{CA}=\cos C\cdot\frac{\sin 2B}{\cos A}\cdot\frac{1}{2\sin B}=\frac{\cos C \cos A}{\cos B}.$
Note that $\displaystyle \frac{HD}{AH}= \frac{\cos C \cos A}{\cos B}$ by direct computation, where $H$ is the orthocenter of $\triangle ABC$ .

This implies $D$ is the orthocenter of $\triangle AEF$ and by using a lemma on the orthocenter and the circumcircle of $\triangle AEF$ , we are done.

This post has been edited 2 times. Last edited by jrpartty, Mar 28, 2022, 6:59 AM
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