Q:

Area of Trapezoid by adding a triangle on top

Let's consider the following trapezoid with added triangle on top: 

I want to derive the formula $A=\frac{a+c}{2}h$ for the area of the trapezoid but I'm not sure how: $$A=\frac{1}{2}a(h+h')-\frac{1}{2}ch'=\frac{1}{2}(ah+ah'-ch'),$$ so I have to show that $$ah'-ch'=ch.$$ My idea was to show $$\frac{a-c}{c}=\frac{h}{h'}$$ using the intercept theorem. But I think I need some help because I don't see how to derive that.

A:

The the top triangle be $\Delta ABE$ clockwise. Let the big triangle be $\Delta ACD$ again in clockwise direction. Let the altitude intersect $\Delta ABE$ at $M$ and $\Delta ACD$ at $N$.
$$\Delta ABE\sim \Delta ACD\;\;\;(\text {AA-similarity})$$
$$\frac {AE}{AD}=\frac {BE}{CD}=\frac ca$$
$$\Delta AME\sim \Delta AND\;\;\;(\text {AA-similarity})$$
$$\frac {AE}{AD}=\frac {AM}{AN}=\frac {h'}{h+h'}$$
Comparing the two results, we arrive at the following-
$$\frac ca=\frac {h'}{h+h'}$$
$$ch+ch′=ah'$$

