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$A'$ is a moving point of side $BC$ of $\triangle ABC$. The perpendicular bisector of $A'B$ and $A'C$ cuts side $AB$ and $AC$ respectively at $B'$ and $C'$. Line $d$ passes through $A'$ and is perpendicular to $B'C'$. Prove that $d$ passes through a fixed point.

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Attempt:

I have predicted that $d$ would pass through point $A''$ in which $AA'' \perp BC$ and $A''$ lies on the circumcircle of $\triangle ABC$. But I haven't found out a way to prove that yet.

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    $\begingroup$ What's the source of the problem? $\endgroup$ – Dr. Mathva Mar 26 '19 at 21:03
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Reflect $A'$ across $B'C'$ to $G$ and let $A'G$ meet circmucircle for triangle $ABC$ at $F$.

  • Since $B'$ is circumcenter for $A'BG$ we have $$\angle BGA' = {1\over 2}\angle BB'A' = \pi/2-\beta$$ and similary $$\angle CGA' = \pi/2-\gamma$$
  • Since $$\angle BGC = (\pi /2-\beta) +(\pi/ 2 -\gamma) = \alpha$$ we see that $G$ is also on circle $(ABC)$

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Finnaly, since $\angle CGA' \equiv \angle CGF$ is fixed so is $F$ and we are done.

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  • $\begingroup$ Why the last equality holds? $\endgroup$ – richrow Jun 12 '19 at 12:33
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    $\begingroup$ Typo: $A\to A'$ @richrow $\endgroup$ – Aqua Jun 12 '19 at 12:34
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Well, when there is no idea then coordinate system comes at handy. And actualy it is an easy problem with c.s.

Let $B=(2b,0)$, $C= (2c,0)$, $A= (0,2a)$ and $A'= (2t,0)$, for some fixed $a,b,c$ and variable $t$. The midpoint of $A'B$ is $N = (b+t,0)$. Since $B'$ is on a line $$AC:\;\;\;{x\over 2b}+{y\over 2c}=1$$ we have $$B' = (b+t,{a(b-t)\over b}) $$ and analougly we get $$C' = (c+t,{a(c-t)\over c}) $$

Now the slope of segment $B'C'$ is $$k= {at\over bc}$$ so the slope of $d$ is $$k' = -{1\over k} = -{bc\over at}$$

So the line $d$ has equation $$ y= {bc\over at}x +{2bc\over a}$$

which means that this line goes always through the point $F=(0,{2bc\over a})$.

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    $\begingroup$ there is a minor sign error while calculating $k'$ it should be $-{bc\over at}$.It's making your answer of opposite sign. $\endgroup$ – Aditya Prakash Mar 26 '19 at 21:50
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    $\begingroup$ Just to add as a fact , if OP wants to do the geometry way , the constant point is the image of triangle's orthocentre in side BC and lies on circumcircle $\endgroup$ – Aditya Prakash Mar 26 '19 at 21:53
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Let $P$ be a point symmetric to $A'$ with respect to line $B'C'$. Firstly, $\angle B'PC'=\angle B'A'C'$ and $\angle B'A'C'=\angle BAC$ (it's easy angle computations). Hence, $\angle B'AC'=\angle B'PC'$, so quadrilateral $B'PAC'$ is cyclic. It means that $\angle PB'B=\angle PC'C$. Also, due to symmetry and properties of perpendicular bissectors $B'P=B'A'=B'B$ and $C'P=C'A'=C'C$, so triangles $BB'P$ and $CC'P$ are isosceles (and similar). Consequently, $\angle PBB'=\angle PC'C$, so quadrilateral $BPAC$ is also cyclic. Let $Q$ be the point on circumcircle of triangle $ABC$ such that $AQ\perp BC$. We will prove that $PQ$ passes through point $A'$. To do that it is sufficient to prove that $$ \frac {S_{PBQ}}{S_{PCQ}}=\frac {BA'}{A'C}. $$ However, $S_{PBQ}=PB\cdot BQ\cdot \sin \angle PBQ$ and similar for $PCQ$. Since $\angle PCQ+\angle PBQ=180^{\circ}$ we obtain $\frac {S_{PBQ}}{S_{PCQ}}=\frac {BP\cdot BQ}{CP\cdot CQ}$. Now note that $\frac {BP}{CP}=\frac {BB'}{CC'}$ and $BA'=2\cos \angle B'BA' \cdot BB'$ and $CA'=2\cos \angle C'CA\cdot CC'$. Thus, it's sufficient to prove that $\frac {BQ}{CQ}=\frac {\cos \angle B}{\cos\angle C}$. But this follows from sine theorem and equalities $\angle BAQ=90^{\circ}-\angle B$ and $\angle CAQ=90^{\circ}-\angle C$.

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