Similarity of triangles

Question

I need help with this geometric problem.

$ \overline {AB}$ is a diameter in circle $M$ , $C \in$ $\overrightarrow{AB}$ lying outside the circle , $\overline{CD}$ is drawn tangent to the circle at point $D$ , then $\overrightarrow{DH} \bot \overrightarrow{AB} $ to intersect it at $H$.

Prove that : $(CD)^2 = CH \times CM = CB \times CA $

What I have proved to far that the $(CD)^2 = CH \times CM $, but I couldn't prove the rest.

Answer 1

Hint. The triangles $CDB$ and $CAD$ are similar: the angle $\angle BDC$ and the angle $\angle DAB$ are equal because they insist upon the same arc $DB$.

Answer 2

The second part can be worked as follows: let $BC=x$ and the radius of the circle be $r$. Then by Pythagoras $(CD)^2$ = $(x+r)^2-r^2$, this simplifies to $2rx+x^2$. Next CB=$x$ and CA=$2r+x$, if you multiply these two you get $2rx+x^2$, the same as $(CD)^2$