If $\angle A-\angle B =90 $. A little angle chasing prove s that $\angle ABH =\angle A-90=B$ .Hence $\Delta HBC$ is $B$-isosceles. Again a little angle chase proves that $\angle HCO=\angle A-\angle B=90 . \Rightarrow RA||CO (R$ is the foot of perpendicular from $C$).But $RH=RC$ From midpoint theorem $RA||CK$ as well. $\Rightarrow C,O,K$ are collinear. If $C,O,K$ are collinear Let $M$ be the midpoint of $BC$ .It is well known that $AH=2.MO$.Let $AM$ intersect $CK$ at $X$.Since $OM ||KA$ and $2.OM=AK,, M$ is the midpoint of $XA$ as well $\Rightarrow ABXC$ is a paralellogram. A little angle chasing shows $\angle BCK =\angle A-90$.But since $AB||CK$ we have $\angle A-90=\angle B$ as desired

Suppose that $C,K,O$ are co-linear. Then $\angle AOC=2B$ and $\angle OAK=\angle C - \angle B \implies \angle AKO = \angle A$. Now, let $M$ be the mid point of $BC$ and $AM$ intersect $CK$ at $N$. It is known that $AH=AK=2OM$ and $AK||MO \implies NM=MA$. Now, as per assumption $MB=MC, \angle AMB=\angle NMC \implies \Delta ABM \cong \Delta NMC$ $ \implies CK||AB \implies \angle AKO + \angle KAB= \angle A + \frac{\pi}{2} - \angle B = \pi$ $\implies \angle A - \angle B= \frac{\pi}{2}$, as desired. Suppose $\angle A - \angle B= \frac{\pi}{2}$. Using trigonometry i.e Law of Sines and Cosines on $\Delta AKC$, find out $\angle AKC$. We would find $\angle AKO= \pi - \angle AKC$.

Another approach through co-ordinate. Take the point $A$ as the origin.From it you can easily find the equations of the perpendicular bisectors and thus you can find the co-ordinates of the orthocentre and the circumcentre.Let the co-ordinate of $B$ be $(x_1,y_1)$ and that of point of $C$ be $(x_2,0)$.Then,little calculation will yield co-ordinate of the orthocentre is $(x_1,\frac{x_1}{tanB})$.As the point $K$ is the reflection of the orthocentre,then the origin is the midpoint of the line segmeny joining the orthocentre and the point $K$.So co-ordinate of the point is $(-x_1,\frac{-x_1}{tanB})$Similarly,you can find the co-ordinate of the orthocentre.Now lies the important part.Determine the equation of the straight line which joins the point $K$ and the vertex$C$. Now as the points $C$,$K$ and $O$ are collinear ,so the co-ordinates of the point $O$ will satisfy the equation.Put the values and you will have the relation $\frac{y_1}{x_2-x_1}=-(\frac{x_1}{y_1})$.Now this relation is eqivalent to $tanB=-\frac{1}{tanA}$ $\rightarrow$$A-B=90$.And for the rest go the reverse way and we are done. I am really desperate to get the other TST problems.

Let $CC'$ be a diameter of the circumcircle ($CABC'$ cyclic). a) If $\hat A=\hat B+90^\circ\implies CC'\parallel AB$ (easy angle chasing). Well known $BC'=AH$; with $AK=AH$ and $AK\bot BC\bot BC'$ we got $ABC'K$ parallelogram, hence $K\in CC'$. b) If $K\in CC'$, with $BC'=AH=AK$ and $AK\parallel BC'$ we got $ABC'K$ a parallelogram, i.e. $AB\parallel CO$, which gives $\hat A=\hat B+90^\circ$, hence we are done. Best regards, sunken rock

Lemma: In $ \triangle ABC $ let $ H, O $ be the orthocenter and circumcenter respectively. Let the reflection of $ H $ wrt $ A $ be $ H' $ and reflection of $ A $ wrt $ M $, the midpoint of $ BC $ be $ D $. Then $ D,O,H' $ are collinear points. Proof: The homothety $ (D,2) $ sends $ M $ to $ A $. Since, $ OM=\frac{HA}{2}=\frac{H'A}{2} $ (well known) and $ OM \parallel HA $, $ O $ and $ H' $ are corresponding points of the homothety and hence $ D,O,H' $ are collinear. In the problem,$ \angle AHC=B, \angle ACH=A-90^{\circ} $. Let $ D $ be the reflection of $ A $ in the midpoint of $ BC $. Then $ K,O,D $ are collinear. So $ KO $ passes thru $ C $ iff $ \angle ACK= 180^{\circ}-A $ i.e, iff $ \angle HCK=90^{\circ} $ i.e, iff $ A $ is the circumcenter of $ \triangle HCK $, i.e, iff $ AH=AC $ i.e, iff $ B=A-90^{\circ} $. Sadly, I bashed it with trigonometry during the test!

Did you comple it?Because I don't think it is too hard using trigo

See PP15 from here.

Solution with vectors: Say $O$ is origin. Then $H=A+B+C$ and then $K = A-B-C$. We see $\angle (A+B,C) = \pi-\alpha+\beta$ and $(A+B)\; \bot \;(A-B)$. Now we have: $K,O,C$ are collinear $\Longleftrightarrow K||C\Longleftrightarrow K = kC \Longleftrightarrow (k+1)C=A-B \Longleftrightarrow $ $C|| A-B \Longleftrightarrow C\bot \; A+B \Longleftrightarrow \pi-\alpha+\beta =\pi/2$. Q.E.D.

If the midpoint of AB is $M$. We know that if $H'$ is the reflection of $H$ of $M$, then $COH'$ is a line. Then note assuming collinearity we have $COH' \equiv COK \equiv KH'$ where the latter is parallel to $AB$. Then we get $180 - A = 90 - B$ so done.

Another approach We will show that $CK+KO=CO$ Use appollonius in $\triangle CKH$ and $\triangle OKH$ to get the lengths $CK$ and $OK$.Now its just trigo....(I think calculations are not too long but boring...)

Apparently this one is an easy complex bash( only doing some labeling is kind of not-so bashy)

Consider a homothety centered at $H$ and with scale $\frac{1}{2}$. $O$ is sent to $N$, the nine-point center of $\Delta ABC$, $C$ goes to $C'$, the midpoint of line segment $CH$. $K$ goes to $A$. Thus $C',A,N$ are collinear. But it is well-known that $C'$ lies on the nine-point circle of $\Delta ABC$, and the midpoint $D$ of $AB$ is its antipode. Thus $A,N,D$ are collinear, that is, $N$ lies on line segment $AB$. This in turn means that $CO \parallel AB$. Consider all angles to be directed. Thus if $\angle DOA = \angle BOD = \theta$, then $\angle AOC = 90^{\circ} - \theta$, and so $$\angle CAB = \angle CAO + \angle OAD = \left ( 45^{\circ} + \frac{\theta}{2} \right ) + (90^{\circ} - \theta) = 135^{\circ} - \frac{\theta}{2},$$and $$\angle ABC = \angle DBO - \angle CBO = 90^{\circ} - \theta - \frac{180^{\circ} - (90^{\circ} + \theta)}{2} = 45^{\circ} - \frac{\theta}{2}$$Thus $$\angle CAB - \angle ABC = 90^{\circ}$$as required.