In $\triangle ABC$ construct altitudes $AA', BB', CC'$ meeting at point M to form the orthocenter of the triangle. Let O be the circumcenter of $\triangle ABC$. Show that $\angle BAA' = \angle OAC$ and $AO \perp B'C'$.
The attempt that I have tried: Extending the altitudes to touch the circle at 3 different points. And also constructing AO and extending it to touch the circle at a point. I attempted to show that arc formed from extending $AA'$ and point B was congruent to the arc formed from extending $AO$ and point $C$. This attempt did not work for me, unless I did something wrong and it actually does work. If someone could please provide guidance for this question I would appreciate it. enter image description here
The first part is easy . . .
Let $\theta = \angle BAA'$.
Then, using degree measure,
\begin{align*} \angle ABA' &= 90 - \theta\\[4pt] \implies \angle ABC &= 90 - \theta&&\text{[since $\angle ABC = \angle ABA'$]}\\[4pt] \implies \angle AOC &= 2(90 - \theta)&&\text{[central angle is twice inscribed angle]}\\[4pt] &= 180 - 2\theta\\[4pt] \implies \angle OAC &= \theta&&\text{[since triangle AOC is isosceles with apex at $O$]}\\[4pt] \end{align*}
Therefore $\angle BAA' = \angle OAC$.
The second part seems harder.
Here's a brute force solution using complex numbers . . .
The goal is to show $OA \perp B_1C_1$.
For each vertex, let the corresponding complex number be denoted by the lower case version of the same letter.
$\qquad\qquad\qquad\qquad$ Without loss of generality, we can assume \begin{align*} &\;\;{\small{\bullet}}\;\;\text{The circumcenter is $0\;$ (by an appropriate translation).} \\[4pt] &\;\;{\small{\bullet}}\;\;\text{The circumradius is $1\;$ (by an appropriate scaling).} \\[4pt] &\;\;{\small{\bullet}}\;\;a = 1\;\text{(by an appropriate rotation).} \end{align*} $\qquad\qquad\qquad\qquad$ Two simple facts:$\;$If $z$ is a complex number, then \begin{align*} &\;\;{\small{\bullet}}\;\;z\;\text{is real if and only if}\;\bar{z}=z \qquad\qquad\qquad\qquad\qquad\qquad\, \\[4pt] &\;\;{\small{\bullet}}\;\;z\;\text{is pure imaginary if and only if}\;\bar{z}=-z \\[12pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Since $|b|=|c|=1$, we have $$ \bar{b}={\small{\frac{1}{b}}}\;\text{and}\;\bar{c}={\small{\frac{1}{c}}} \qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\, $$
$\qquad\qquad\qquad\qquad$ Since $B_1$ is on line $AC$,
\begin{align*} b_1 &= 1 + s(c-1),\;\text{for some}\;s \in\mathbb{R} \qquad\qquad\qquad\;\;\;\;\; \\[4pt] \implies\;\overline{b_1} &= 1 + s(\bar{c} - 1) \\[4pt] &=1 + s\left({\small{\frac{1}{c}}} - 1\right) \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Next, let
$$ p = \frac{b - b_1}{c-1} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\; $$
$\qquad\qquad\qquad\qquad$ Then
\begin{align*} &B_1B \perp AC \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\, \\[4pt] \implies\; &p\;\text{is pure imaginary} \\[4pt] \implies\; &\bar{p}=-p \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Simplifying $p$, we get \begin{align*} p &= \frac{b - b_1}{c-1} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\; \\[4pt] &= \frac{b- \left(1 + s(c-1)\right)}{c-1} \\[4pt] &= \frac{b-1}{c-1} - s \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Then
\begin{align*} p &= \frac{b-1}{c-1} - s \qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;\; \\[4pt] \implies\;\bar{p} &= \frac{\bar{b}-1}{\bar{c}-1} - s \\[4pt] &= \frac{\frac{1}{b}-1}{\frac{1}{c}-1} - s \\[4pt] &= \frac{c(b-1)}{b(c-1)} - s \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Hence
\begin{align*} &\bar{p}=-p \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; \\[6pt] \implies\; &\frac{c(b-1)}{b(c-1)} - s = s - \frac{b-1}{c-1} \\[4pt] \implies\; &s = \frac{(b+c)(b-1)}{2b(c-1)} \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Then
\begin{align*} b_1 &= 1 + s(c-1) \qquad\qquad\qquad\qquad\qquad\;\;\;\;\; \\[4pt] &= 1 + \left(\frac{(b+c)(b-1)}{2b(c-1)}\right)(c-1) \\[4pt] &= 1 + \frac{(b+c)(b-1)}{2b} \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Thus, we have
$$ b_1 = 1 + \frac{(b+c)(b-1)}{2b} \qquad\qquad\qquad\qquad\;\;\;\; $$
$\qquad\qquad\qquad\qquad$ By analogous reasoning, we get
$$ c_1 = 1 + \frac{(b+c)(c-1)}{2c} \qquad\qquad\qquad\qquad\;\;\;\; $$
$\qquad\qquad\qquad\qquad$ Then
\begin{align*} &OA \perp B_1C_1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\; \\[6pt] \iff &\frac{c_1-b_1}{a}\;\text{is pure imaginary} \\[4pt] \iff\; &c_1-b_1\;\text{is pure imaginary}\qquad\text{[since a = 1]} \\[4pt] \end{align*}
$\qquad\qquad\qquad\qquad$ Let $w = c_1-b_1.\,$ It remains to show $\bar{w} = - w.\;$Then
\begin{align*} w &= c_1-b_1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \\[4pt] &= \left(1 + \frac{(b+c)(c-1)}{2c}\right)-\left(1 + \frac{(b+c)(b-1)}{2b}\right) \\[4pt] &= \frac{c^2-b^2}{2bc} \\[4pt] \implies\; \bar{w} &= \frac {\left(\frac{1}{\large{c}}\right)^2 - \left(\frac{1}{b}\right)^2} { 2 \left(\frac{1}{b}\right) \left(\frac{1}{{\large{c}}}\right) } \\[4pt] &= \frac{b^2-c^2}{2bc} \\[4pt] \implies\;\bar{w} &= - w,\;\text{as required} \\[4pt] \end{align*}
This completes the proof.