A right angle relates to orthocenter and circumcenter

Question

I'm struggling to solve this. It would be very nice if you use no knowledge on circle to solve this (this problem was introduced before the context of circle).

Let $ABC$ be an acute triangle, with orthocenter $H$, circumcenter $O$. The line parallel to BC passing through $O$ intersects AC at K. Let $I$ be the midpoint of $AH$. Prove that $\widehat{BIK}=90^o$.

My attempts:
I drew three altitudes, and called them $AD$, $BE$, $CF$ respectively. Let $J$ be the intersection of $EF$ and $AH$. Then I can see that $CJ$ is perpendicular to $BI$ and I'm trying to prove that $CJ$ and $IK$ are parallel.

Please help me. Thanks.

Answer 1

Let's place your problem on the coordinate system. We may assume that: $$A=(a,b), \\ B=(-1,0), \\ C=(1,0).$$ Therefore, the equation of $AC$ is $y=\frac{b}{a-1}(x-1).$ And, $O$ is the intersection of $x=0$ and the perpendicular bisector of $AC$ whose equation is $y-\frac{b}{2}=(\frac{1-a}{b})(x-(\frac{a+1}{2})).$ Thus,

$$O=(0, \frac{a^2+b^2-1}{2b}).$$

Moreover, $K$ is the intersection of $y=\frac{a^2+b^2-1}{2b}$ and $AC$. So,

$$K=(\frac{a^3+ab^2-a-a^2+b^2+1}{2b^2}, \frac{a^2+b^2-1}{2b}).$$

On the other hand, $H$ is the intersection of $x=a$ and the altitude of the triangle passing through $B$ whose equation is $y=(\frac{1-a}{b})(x+1)$. Hence,

$$H=(a, \frac{1-a^2}{b}), $$

and,

$$I=(a, \frac{1-a^2+b^2}{2b}). $$

Now, if $m$ and $m'$ are respectively the slopes of $BI$ and $IK$, we just need to show $mm'=-1$, while we have:

$$mm'=\frac{\frac{1-a^2+b^2}{2b}}{a+1} \times \frac{\frac{a^2+b^2-1}{2b}-\frac{1-a^2+b^2}{2b}}{\frac{a^3+ab^2-a-a^2+b^2+1}{2b^2}-a} \\ =\frac{\frac{1-a^2+b^2}{2b}}{a+1} \times \frac{\frac{a^2-1}{b}}{\frac{a^3-ab^2-a-a^2+b^2+1}{2b^2}} \\ = \frac{(1-a^2+b^2)(a-1)}{a^3-ab^2-a-a^2+b^2+1}=-1.$$

we are done.

Answer 2

There are an extreme amount of facts in this configuration. Letting $E$ be such that $OKCE$ is an isosceles trapezoid, and $D$ being the reflection of the orthocenter of $BC$, we get many facts. Let $M=\frac{B+K}{2}$. My favorites are i) $OKEB$ is a parallelogram, and ii) $DE=EO=OK=KC$ and $D,O,E,K,C$ lie on a circle.

To actually solve the problem, there are many approaches. Here's one that I thought was not too hard to find:

Draw the circle centered at $M$ passing through $B,K$, and call it $\Omega$. Then $D\in \Omega$. This isn't hard to prove; show that $OM$ is the perpendicular bisector of $BD$, and hence $MB=MD$. One way to show that $OM$ is the perpendicular bisector is with $E$. I claim $OKEB$ is a parallelogram, so that $M$ is also the midpoint of $OE$. This is true because $OK||EB$ and $\angle OKB=\angle OCB=\angle OBC$. Since $OB=OD$ is the radius $R$ of $(ABC)$, it's enough to show $OE$ is said bisector. Look at angles. It's enough to show $D,O,E,K$ are concyclic.

You can finish by showing $I\in \Omega$ via angles (which solves the problem by Thales), or by directly showing $\angle BIK=\pi/2$ via lengths -- $IB^2+IK^2=DB^2+DK^2$. To compute $IB$, note that it's a median in $\triangle BHA$. To compute $DB$, note $DB=BH$. To compute $IK$, look at $\triangle IAK$ and observe $KO=KC=EB$. Finally, to compute $DK$, note $DEKC$ is a cyclic isosceles trapezoid, so $KD=EC$.

The details are not difficult to fill in. The below diagram is enlightening: