# Prove that $d$ passes through a fixed point.

$$A'$$ 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.

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.

• What's the source of the problem? – Dr. Mathva Mar 26 '19 at 21:03

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 $$(ABC)$$

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

• Why the last equality holds? – richrow Jun 12 '19 at 12:33
• Typo: $A\to A'$ @richrow – Aqua Jun 12 '19 at 12:34

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 $$F=(0,{2bc\over a})$$.

• there is a minor sign error while calculating $k'$ it should be $-{bc\over at}$.It's making your answer of opposite sign. – Aditya Prakash Mar 26 '19 at 21:50
• 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 – Aditya Prakash Mar 26 '19 at 21:53

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 $$\angle CAQ=90^{\circ}-\angle C$$.