analytic geometryeuclidean-geometrygeometry
$A'$ 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.
Ignoring vertex $A$, this becomes a problem on the inscriptable $\square BCNM$. Let $L$ be the midpoint of $\overline{BC}$. Also, let $M'$ (instead of $P$), $N'$ (instead of $Q$), $H'$, $K'$ be the projections of $M$, $N$, $H$, $K$ onto $\overline{BC}$. Let the tangent segments from $B$ and $C$ to the incircle have length $d$; let the tangent segments from $M$ and $N$ have length $m$ and $n$. Finally, define $m' := |MM'|$, $n':=|NN'|$, $m'':=|M'L|$, $n'':=|N'L|$. Without loss of generality, $m\leq n$ so that $m'\leq n'$.
Certainly, if $m=n$, then $\overleftrightarrow{HK}$ meets $\overline{BC}$ at $L$. For $m \neq n$, we'll prove that $L$ is on $\overleftrightarrow{HK}$ by showing $\triangle HH'L\sim \triangle KK'L$ via $$|HH'||K'L|=|KK'||H'L| \tag{$\star$}$$
Parallelism and proportionality rules tell us that $$\frac{|M'H'|}{|M'N'|}=\frac{|MH|}{|MN|}=\frac{m}{m+n} \qquad |HH'|=m'+\frac{m}{m+n}(n'-m')=\frac{m'n+mn'}{m+n} \tag{1}$$ The Crossed Ladders Theorem tells us that $$\frac{1}{|KK'|}=\frac{1}{m'}+\frac{1}{n'} \quad\to\quad |KK'| = \frac{m'n'}{m'+n'} \tag{2}$$ (and, in fact, $K$ is the midpoint of the extension of $\overline{KK'}$ that meets $\overline{MN}$), whereupon some proportional thinking then yields $|M'K'|:|K'N'|=m':n'$, so that we have $$\frac{|M'K'|}{|M'N'|}=\frac{m'}{m'+n'} \tag{3}$$ Therefore, $$\begin{align} |H'L|&=|M'L|-|M'H'| = m'' - \,\frac{m}{m+n}(m''+n'') \;= \frac{m''n-mn''}{m+n} \\[6pt] |K'L|&=|M'L|-|M'K'| = m'' - \frac{m'}{m'+n'}(m''+n'')=\frac{m''n'-m'n''}{m'+n'} \end{align} \tag{4}$$ Substituting in $(\star)$, and clearing denominators, we need only verify that $$(m'n+mn')(m''n'-m'n'') = m'n'(m''n-mn'') \tag{5} $$ That is, $$\frac{m}{n}\cdot\frac{m''}{n''} = \left(\frac{m'}{n'}\right)^2 \tag{6}$$
It seems like there's a geometric mean argument to be made, but I'm not seeing it. So, writing $\theta$ for the common angle at $B$ and $C$, we have $$\frac{m}{n}\cdot\frac{d-(m+d)\cos\theta}{d-(n+d)\cos\theta} = \left(\frac{m+d}{n+d}\right)^2 \quad\to\quad (d+m)(d+n)\cos\theta = d^2 - m n \tag{7}$$ This same relation results (for $\theta \neq 0$) from the observation that there's a right triangle with hypotenuse $|MN|$ and legs $|m'-n'|$ and $m''+n''$. $$(m+n)^2 = (m'-n')^2 + (m''+n'')^2 \qquad\to\qquad (7) \tag{8}$$ This equality establishes $(\star)$ and completes the proof. $\square$
I believe there's a cleaner way to link $(6)$ and $(8)$ (or to demonstrate $(6)$ some other way) without having to show equality through $(7)$. Again, I'm not seeing it. Perhaps I'll return to this question.
Best Answer
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 $F=(0,{2bc\over a})$.