$ABCD: AB ||CD, AB>CD, AD=BC$
$k(O)$ inscribed
$DH \bot AB,H \in AB$ and $\angle ADC = \gamma$
$\angle BHC, \angle BCH =$ ?
I have tried to show that $\triangle BCH$ is isosceles, but when I made the diagram, I noticed that it isn't. $\angle ADC = \angle DCB = \gamma$, thus $\angle DAB = \angle ABC = 180^\circ - \gamma$. I'm not sure what to do now.
On
One of the properties of tangential trapezoid is that sum of the bases is equal to sum of the legs. See Pitot theorem. So $AB+CD=2BC$, $\frac{AB}{2}+\frac{CD}{2}=BC$. But it's easy to see that $\frac{AB}{2}+\frac{CD}{2}=HB$.
The lengths of the tangents from one point to a circle are equal (due to symmetry you have congruent triangles). Let's call $A'$ the point where the circle is tangent to $AB$, $B'$ the point where the circle is tangent to $BC$ and so on. Then $$A'B=\frac 12 AB=BB'$$ Similarly $$B'C=CC'=\frac 12 CD=C'D$$ Since $A'C'$ is perpendicular to both $AB$ and $CD$, and $AB||CD$, and $DH\perp AB$, you got that $HA'C'D$ is a rectangle, so $$HA'=C'D$$ From these equations $$HB=HA'+A'B=C'D+A'B=B'C+BB'=BC$$ Therefore $\triangle HBC$ is isosceles ($HB=BC$), and $\angle BHC=\angle BCH$.