Find $|CM|$, if $|CA|=a$ and $|CB|=b$.

Question

Let $O$ be a center of a circle, circumscribed over $\triangle ABC$. Perpendicular, drown from the point $A$ on the line $CO$, cross the line $CB$ in the point $M$. Find $|CM|$, if $|CA|=a$ and $|CB|=b$.

Answer 1

You have $AC=CD=a$, because $CO\perp AM$. Also $\angle HCD=90-\beta$

$\frac{CH}{CM}=\cos(90-\beta+\beta-\alpha)=\cos (90-\alpha)=\sin\alpha \quad\quad(1)$.

Also $\frac{CH}{a}=\sin\beta\quad\quad (2)$

From $(1)$ and $(2)\Rightarrow CM=\frac{CH}{\sin\alpha}=\frac{a\sin\beta}{\sin\alpha}=\frac{a^2}{b}$, because from the $\sin$-theorem we have $\frac{a}{\sin\beta}=\frac{b}{\sin\alpha}$