Suppose a trapezium $ABCD$. There are circles $m,n$ with centres midpoint of leg $BC=M_{1}$ and leg $AD=M_{2}$, respectively; and diametres $BC$ and $AD$, respectively. The point $P$ is the intersection of $BC$ and $AD$. We have tangent lines to $m,n$ from $P$. Prove that the angle $\alpha$ between the tangents to $m$ is equal to $\beta$ between the tangents to $n$.
Since the tangents touch the circle only at one point, they are perpendicular to the radii. Therefore, $$\sin\Bigl(\frac{\alpha}{2} \Bigr)=\frac{r}{|M_{1}P|}\implies \alpha=2\sin^{-1}\left(\frac{|CB|}{2|M_{1}P|} \right)~.$$ Likewise for $\beta$.
It is then enough to prove that $\displaystyle\frac{|CB|}{|M_{1}P|}=\frac{|AD|}{|M_{2}P|}$.
I have had trouble with proving that last part. Your help would be really appreciated!
Best Answer
Say the perpendicular distance from point $P$ to the base of trapezium is $h$ and it meets the base at $H$. Also, perpendicular from the midpoint of the leg $M_1$ meets the base at $H_1$ and say the perp distance from the midpoint to the base is $d$. (Note: given $AB \parallel CD, $ perp distance from $M_2$ to the base of the trapezium will also be $d$ using midpoint theorem)
First note that $\triangle PHB \sim \triangle M_1H_1B$
So, $ ~ \displaystyle \frac{r_1}{d} = \frac{PM_1 + r_1}{h} \implies PM_1 = \frac{r_1 (h - d)}{d}$
where $r_1$ is radius of the circle on leg $BC$.
Now using the right triangle formed by $P, M_1$ and the point of tangency to circle $m$,
$ \displaystyle \sin \frac{\alpha}{2} = \frac{r_1}{PM_1} = \frac{d}{h-d}$
Similarly show that $ \displaystyle \sin \frac{\beta}{2} = \frac{d}{h-d}$
That leads to $\alpha = \beta$.