Here is a TikZ diagram to accompany the argument provided.
\begin{tikzpicture}
%Trapezoid ABCD is drawn. The lines through its legs intersect at P. The intersection
%of the diagonals is labeled O. The intersection of the line through O and P
%intersects the bases at M and N.
\coordinate (A) at (0,0);
\node[anchor=north, inner sep=0, font=\footnotesize] at (0,-0.15){$A$};
\coordinate (B) at (6,0);
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) +(0,-0.15)$){$B$};
\coordinate (C) at ($(B) +(-2.5,4)$);
%\node[anchor=north, inner sep=0, font=\footnotesize] at ($(C) +(0,-0.15)$){$C$};
\coordinate (D) at ($(A) +(1,4)$);
%\node[anchor=north, inner sep=0, font=\footnotesize] at ($(D) +(0,-0.15)$){$D$};
%
\draw (A) -- (B) -- (C) -- (D) -- cycle;
%
%Lines through the legs of the trapezoid are drawn.
%
\path[name path=a_path_to_locate_P] let \p1=($(A)-(D)$), \n1={atan(\y1/\x1)} in (D) -- ($(D) +(\n1:3.25)$);
\path[name path=another_path_to_locate_P] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)} in (C) -- ($(C) +({\n1+180}:3.75)$);
%
\coordinate[name intersections={of=a_path_to_locate_P and another_path_to_locate_P, by=P}];
%
\draw[dashed] (P) -- (C);
\draw[dashed] (P) -- (D);
%
%
%The labels for the vertices of the trapezoid are typeset.
\path[name path=a_path_to_typeset_C] ($(B)!0.15cm!-90:(P)$) -- ($(P)!0.15cm!90:(B)$);
\path[name path=another_path_to_typeset_C] (C) -- ($(C)!-0.2cm!(D)$);
\coordinate[name intersections={of=a_path_to_typeset_C and another_path_to_typeset_C, by=label_C}];
\node[anchor=west, inner sep=0, font=\footnotesize] at (label_C){$C$};
\path[name path=a_path_to_typeset_D] ($(A)!0.15cm!90:(P)$) -- ($(P)!0.15cm!-90:(A)$);
\path[name path=another_path_to_typeset_D] (D) -- ($(D)!-0.2cm!(C)$);
\coordinate[name intersections={of=a_path_to_typeset_D and another_path_to_typeset_D, by=label_D}];
\node[anchor=east, inner sep=0, font=\footnotesize] at (label_D){$D$};
%
%
%The label for P is typeset.
\coordinate (M) at ($(A)!0.5!(B)$);
\coordinate (N) at ($(C)!0.5!(D)$);
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=\n1, inner sep=0, font=\footnotesize] at ($(P) +({\n1-180}:0.15)$){$P$};
%The line segment from P through M and N is drawn.
\draw[dashed] (P) -- (M);
%The diagonals of the trapezoid are drawn.
\draw[name path=a_path_to_locate_O, dashed] (A) -- (C);
\draw[name path=another_path_to_locate_O, dashed] (B) -- (D);
%
\coordinate[name intersections={of=a_path_to_locate_O and another_path_to_locate_O, by=O}];
%Labels for the midpoints of the bases of the trapezoid are typeset.
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=south west, inner sep=0, font=\footnotesize] at ($(M) +({0.5*(\n1+180)}:0.15)$){\textit{M}};
\draw let \p1=($(M)-(N)$), \n1={atan(\y1/\x1)} in node[anchor=south west, inner sep=0, font=\footnotesize] at ($(N) +({0.5*(\n1+180)}:0.15)$){\textit{N}};
%A "pin" is drawn to M and N.
\draw[draw=gray, shorten <=1mm, shorten >=1mm] (O) -- ($(O) +(0.5,0)$);
\node[anchor=west, inner sep=0, font=\footnotesize] at ($(O) +(0.5,0)$){\textit{O}};
\end{tikzpicture}
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, $XY = XO +YO =9$.
Best Answer
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.