Let ABCD be a trapezoid, such that AB is parallel to CD. Let $ABCD$ be a trapezoid, such that $AB$ is parallel to $CD$. Through $O$, the intersection point of the diagonals $AC$ and $BD$ consider a parallel line to the bases. This line meets $AD$ at $M$ and $BC$ at $N$. 
Prove that $OM=ON$ and: $$\frac{2}{MN}=\frac1{AB}+\frac1{CD}$$
 A: First problem: We use an area argument.  Draw the trapezoid, with $A,B,C,D$ going counterclockwise, and $AB$ a horizontal line at the "bottom." (We are doing this so we will both be looking at the same picture.)
Note that $\triangle ABC$ and $\triangle ABD$ have the same area. (Same base $AB$, same height, the height $h$ of the trapezoid.)
These two triangles have $\triangle ABO$ in common. It follows that $\triangle OBC$ and $\triangle OAD$ have the same area.
Let $h_1$ be the perpendicular distance from $AB$ to $MN$, and $h_2$ the perpendicular distance from $MN$ to $DC$. 
The area of $\triangle OBC$ is $\frac{1}{2}(ON)(h_1+h_2)$. This is because it can be decomposed into $\triangle OBN$ plus $\triangle ONC$. These have bases $ON$, and heights $h_1$ and $h_2$.
Similarly, $\triangle OAD$ has area $\frac{1}{2}(OM)(h_1+h_2)$.
By cancellation, $ON=OM$. 
Second problem: Here we will use similar triangles. Let $h$, $h_1$, and $h_2$ be as in the first problem. By using the first problem, and the fact that triangles $ACD$ and $AOM$ are similar, we get
$$\frac{CD}{MN/2}=\frac{h}{h_1}.$$
A similar argument shows that
$$\frac{AB}{MN/2}=\frac{h}{h_2}.$$
Invert. We get
$$\frac{MN/2}{CD}=\frac{h_1}{h}\quad\text{and}\quad \frac{MN/2}{AB}=\frac{h_2}{h}.$$
Add, and use the fact that $h_1+h_2=h$. We get 
$$\frac{MN/2}{CD}+\frac{MN/2}{AB}=1.$$
This yields 
$$\frac{MN}{2}=\frac{(AB)(CD)}{AB+CD}.$$
Invert both sides. We get the desired result. 
A: 
The picture support the above solution.
On the other hand ot is possible to go another way.
Trinagles AMO and ACD are similar. Then $\frac{MO}{DC}=\frac{AO}{AC}$
Trinagles BMO and BCD are similar. Then $\frac{NO}{DC}=\frac{BO}{BD}$
Trinagles DOC  and AOB are similar. Then $\frac{DO}{OB}=\frac{OC}{AO}$
From the last equality it follows that  $\frac{DB}{OB}=\frac{AC}{AO}$
Then  $\frac{MO}{DC}=\frac{NO}{DC}$ So $MO=NO$
