it is just Click to reveal hidden text

to tackle this problem.

I don't know about this, but I have \[ \angle ABP=\angle PBQ=\angle QBC, \] = 8.948922449° (yes, that ugly angle) and $ \frac{BP}{CH} $ = 2.

Yes, the angle measurement of 8.948922449.....° is correct.

Hint: When the sum of two angles equals 90 degrees, the sum of the squares of the sine values of the angles equals 1.

by a simple trigonometric calculation we get $\angle A=63.15323236...,\angle B=26.84676764...$ whose sum is 90.

I don't know how valid is this argument but its probably the same as trig ceva. From $\angle PBA = \angle QBC$ and $\angle PAC = \angle QAB$ we have that $P$ and $Q$ are isogonal conjugates. Therefore , $CM$ is the reflection of $CH$ over the bisector of $\angle ACB$ and therefore we get that the circumcenter of $ABC$ lies in $CM$. Therefore we must have $\angle ACB = 90$. This concludes part (a). Now let $\angle CAB = \alpha$. By easy angle chase we get $\angle ABP = 30 - \frac{\alpha}{3}$, $\angle PBC = \angle 60 - \frac{3\alpha}{2}$. We have $BP = \frac {BP}{\sin (60 + \frac{\alpha}{3})} =\frac{BC \cdot \sin \alpha}{\sin (60 + \frac{\alpha}{3})}$ and $CH = BC \cdot \cos \alpha$ and therefore we obtain \[\frac{BP}{CH} = \frac{\sin \alpha}{\cos \alpha \cdot \sin 60 + \frac{\alpha}{3}} (*)\] From trig ceva for cevians $AP,CP,BP$ we get $\sin \alpha = \frac{ \cos \alpha \cdot \sin 60 - \frac{2\alpha}{3}}{\sin 30 - \frac{\alpha}{3}}$ so plugging this in $(*)$ gives us \[ \frac{BP}{CH} = \frac{\sin 60 - \frac{2\alpha}{3}}{ \sin 30 - \frac{\alpha}{3} \cdot \sin 60 + \frac{\alpha}{3}} = \frac{\sin 60 - \frac{2\alpha}{3}}{ \sin 30 - \frac{\alpha}{3} \cdot \cos 30 - \frac{\alpha}{3}} = 2 \]

Here's a synthetic solution to (b). Reflect $P$ about $AB$ to $P'$. Then $\angle CBP = 2 \angle PBH = \angle PBP'$, so by similar triangles and two applications of Angle Bisector Theorem we have \[\frac{CB}{CH} = \frac{AC}{AH} = \frac{CP}{PH} = 2\frac{CP}{PP'} = 2\frac{CB}{P'B} = 2\frac{CB}{PB},\]thus $\frac{BP}{CH} = 2$.

Anyone can proof a? Since i dont know how to use isogonal conjaquates