Find the measure of the arc $HO$.

Question

There is an acute triangle $ABC$. Take the vertex $B$ as the center and draw a circle arc that contains $H$ and $O$. If the measure of the $\angle BAC$ equals $50^\circ$, calculate the measure of the arc $HO$. ($H$ and $O$ are the orthocenters and circumcenter of the triangle $ABC$). (Answer: $20$ degrees)

$ACG= 180^\circ - 50^\circ-90^\circ =40^\circ\\ \angle BOC =50^\circ\cdot2 = 100^\circ\\ \angle OBC = \angle OCB = 40^\circ$

Answer 1

Assuming $BO=BH$, draw circle with radius $BO$, intersecting circumcircle at $D$. Since the circles are equal, triangle $DOB$ is equilateral and $\angle BOD=60^o$. And since $\angle BOC=2\angle BAC=100^o$, then $\angle DOC=100^o-60^o=40^o$, and $\angle DBC=\frac{1}{2}\angle DOC=20^o$.

And since $AD$ through orthocenter $H$ is perpendicular to $BC$, then $BC$ bisects chord $HD$, making $\angle CBH=\angle DBC=20^o$.

Therefore, $\angle HBO=60^o-40^o=20^o$.

Answer 2

Hint: consider following facts:

1-B is on the perpendicular bisector of HO .

2 the radius of the arc HO is equal to the radius of circumcircle $R$.

4- Find the radius of circumcircle $R$ and inscribed circle $r$.

3- mark center of inscribed circle as I, calculate $(OI=d)^2=R(R-2r)$

4- $HO\approx 2d$.

Now you have the measure of the chord HO and radius of arc HO, using them you can find the measure of arc HO.

Note: I think you need the measures of sides of triangle to find R and r. There may be different solution without these measures.