$ABCD$ is a trapezoid $(AB \parallel CD; AB=a;CD=b;a\ne b)$. The circumscribed circle $k(O; R)$ of $\triangle BCD$ is tangent to $AD$ at $D$. Can we find $BD$?
I was solving a similar problem, and now I am curious if we can find $BD$ using only the bases of the trapezoid and knowing $k$ is tangent to $AD$ at $D$. I am thinking about similar triangles but do not happen to find them here. Can you give me a hint?
As per Alternate Segment theorem, $\angle ADB=\angle BCD$. Also, $\angle DBA=\angle CDB$ (angles in parallel lines). Thus, $\triangle ADB\sim\triangle BCD$ and so $AB:BD=BD:DC$, i.e. $AB\cdot CD=BD^2$, i.e. $BD=\sqrt{ab}$.