A height in isosceles triangle isosceles triangle with height and distances
$$M \in  AB$$
$$MM_1 \bot  AC, M_1 \in AC$$
$$MM_2 \bot  BC, M_2 \in BC$$
$$AH \bot  BC, H \in BC$$
I have to show that: $$MM_1 + MM_2 = AH.$$
I've forgotten how this happens and I hope you will help me. :) 
 A: Because $$\frac{AC\cdot MM_1}{2}+\frac{BC\cdot MM_2}{2}=\frac{BC\cdot AH}{2}$$ and since $AC=BC,$ we are done!
A: 
Draw the altitude $BH_2$, so that $BH_2 \bot AC, \; H_2\in AC$.
By Thales' Theorem
$$\frac{AH}{MM_2}=\frac{AB}{MB}\iff \frac{AH}{AB}=\frac{MM_2}{MB} \quad \quad \quad \quad \quad \;\;(1)$$
$$\frac{BH_2}{MM_1}=\frac{AH}{MM_1}=\frac{AB}{AM}\iff \frac{AH}{AB}=\frac{MM_1}{AM} \quad \quad (2)$$
Combining (1) and (2)

$$\frac{AH}{AB}=\frac{MM_2}{MB}=\frac{MM_1}{AM}=\frac{MM_2+MM_1}{MB+AM}=\frac{MM_2+MM_1}{AB}$$ $$\therefore AH=MM_1+MM_2$$


Alternatively

Lemma
The sum of all angles in a triangle is $180°$.


In this solution, you'll have to take a point $P\in AH$, such that $MP \bot AH$. Make sure you now understand why $PMM_2H$ is a rectangle, which implies, that $\angle APM=90°$ and that $\color{red}{PH=MM_2}$. We will now show $$\Delta AMM_1 \cong \Delta AMP$$ i.e., triangles $\Delta AMM_1$ and $\Delta AMP$ are congruent.
In the triangle $\Delta ABH$, let $\angle BAH=\alpha$ and $\angle HBA=\beta$. Hence (due to the angle-sum in triangles) $$\angle BAH+\angle AHB+ \angle HBA= \alpha +90°+\beta =180°$$
In the triangle $\Delta AMP$ $$\angle MAP +\angle APM+\angle PMA=\alpha +90°+\angle PMA=180°\iff \angle PMA=\beta$$
Note furthermore, that $\angle HBA=\angle CBA=\angle BAC=\angle MAM_1$ since the triangle is isosceles.
This implies $\angle MAM_1=\beta$
Now, regarding the triangle $\Delta AMM_1$ $$\angle MAM_1+\angle AM_1M+\angle M_1MA=\beta +90°+\angle M_1MA=180°\iff \angle M_1MA=\alpha$$
Since triangles $\Delta AMP$ and $\Delta AMM_1$ have equal angles and share the side $AM$, they are congruent, which implies $\color{red}{AP=MM_1}$
Taken the red highlighted results

$$AH=PH+AP=MM_2+MM_1$$

Q.E.D.
