A problem about arc, angle in circles Let $M$ is the midpoint of the arc ${AB}$ of the circle $(C)$, $E$ is a point of the arc $MB$. Let $H$ is the perpendicular projection of $M$ on $AE$. Prove that: $AH=HE+EB$.
i tried to draw more line but get stuck on HE, they just said E is an point on the arc MB. 
Can anyone show me some ideas for this proof? this is a problem when i study about arc. Thank you for answer.
 A: Let $\measuredangle ACM=\measuredangle MCB=\alpha$ and $\measuredangle MEC=\beta.$
Thus, $$AH=AM\cos\measuredangle MAE=2R\sin\frac{\alpha}{2}\cos\frac{\beta}{2},$$
$$EH=EM\cos\measuredangle MEA=2R\sin\frac{\beta}{2}\cos\frac{\alpha}{2}$$ and
$$EB=2R\sin\frac{\alpha-\beta}{2}.$$
Id est, we need to prove that:
$$\sin\frac{\alpha}{2}\cos\frac{\beta}{2}=\sin\frac{\beta}{2}\cos\frac{\alpha}{2}+\sin\frac{\alpha-\beta}{2}.$$
Can you end it now?
A:  
Let $\angle AMO = \alpha$, $\angle MOE = \beta $ and $r$ be the radius of the circle. Then,
$$AH = AM\cos\frac{\beta}2 = 2r\cos\alpha\cos\frac{\beta}2$$
Note that $y=180-2\alpha$. Then,
$$HE = AE - AH  = 2r\sin\frac{y+\beta}2 - AH
=2r\cos(\alpha - \frac{\beta}2)- 2r\cos\alpha\cos\frac{\beta}2=2r\sin\alpha\sin\frac{\beta}2$$
Also, note that $x = 180 - 2\alpha - \beta$. Then,
$$EB = 2r \sin\frac x2 = 2r\cos(\alpha + \frac{\beta}2)$$
Therefore,
$$HE + EB = 2r\sin\alpha\sin\frac{\beta}2 + 2r\cos(\alpha + \frac{\beta}2)= 2r\cos\alpha\cos\frac{\beta}2=AH$$
A: Consider the figure below, where $AE$ has been produced to $B'$, so that $EB \cong EB'$.



*

*Observe that $\angle MEB$ and $\angle MAB$ are supplementary. Moreover, $\angle MEA\cong\angle MBA\cong \angle MAB$.

*By means of 1., demonstrate that $\triangle MEB \cong \triangle MEB'$ (SAS criterion).

*Use 2. to show that $\triangle MAB'$ is isosceles, implying $AH \cong HB'$. And the thesis follows immediately.

