A problem about arc, angle in circles

Question

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.

Answer 1

Consider the figure below, where $AE$ has been produced to $B'$, so that $EB \cong EB'$.

1. Observe that $\angle MEB$ and $\angle MAB$ are supplementary. Moreover, $\angle MEA\cong\angle MBA\cong \angle MAB$.
2. By means of 1., demonstrate that $\triangle MEB \cong \triangle MEB'$ (SAS criterion).
3. Use 2. to show that $\triangle MAB'$ is isosceles, implying $AH \cong HB'$. And the thesis follows immediately.

Answer 2

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,

$$HE + EB = 2r\sin\alpha\sin\frac{\beta}2 + 2r\cos(\alpha + \frac{\beta}2)= 2r\cos\alpha\cos\frac{\beta}2=AH$$

Answer 3

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?