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.
(1) After constructing the dotted parallel line as shown, we have two //grams with sides of length = 437.
(2) The length of the base of the larger triangle = 138
(3) The length of the base of the smaller triangle = 138* (3/5) = x, say.
(4) The required = x + 437.
Best Answer
Let AC and BD cross at O. Then, the similar triangles lead to $\frac{XO}{AB}= \frac{XD}{DA},\> \frac{XO}{DC}= \frac{XA}{AD}$. Add up the two ratios to get
$$\frac{XO}{AB}+ \frac{XO}{DC}=1 $$ which yields $XO = \frac{AB\cdot DC}{AB+DC}=\frac92$. Likewise, $YO= \frac92$. Thus, $XY = XO +YO =9$.