How to prove $XP = X'P$? Two triangles $\triangle ABC$, $\triangle A'BC$ have the same base and the same height. Through the point $P$ where their sides intersect we draw a straight line parallel to the base; this line meets the other side of $\triangle ABC$ in X and the other side of $\triangle A'BC$ in $X'$. Prove that $XP = X'P$.
 A: Observe that $\triangle XPA \sim \triangle ABC$ and $\triangle X'PA' \sim \triangle A'BC$. As the $XX'$ line is parallel to $BC$, and $\triangle ABC$ and $\triangle A'BC$ were of the same height, then $\triangle XPA$ and $\triangle X'PA'$ are also of the same height. The result follows from the fact, that the bases have the same length, that is $|XP| = |X'P|$ because $|BC| = |BC|$.
I hope this helps ;-)
Edit: Solution using trapezium area (as asked for in the comments):
We start with the area for trapezium
$$|ABCA'| = |BCP| + |A'AP| + |BXP| + |CX'P| + |XPA| + |X'PA'|$$
then relate the areas of two triangles of the same height
\begin{align}
|BCP| + |BXP| + |XPA| + |A'AP| &= |ABC| \\
= |A'BC| &= |BCP| + |CX'P| + |X'PA'| + |A'AP|
\end{align}
and combining both we get
$$|BXP| + |XPA| = |CX'P| + |X'PA'|.$$
However $\triangle BXP$ and $\triangle CX'P$ are of the same height as well as $\triangle XPA$ and $\triangle X'PA'$. Writing this down we obtain
$$\frac{1}{2}b_1*(h_1+h_2) = \frac{1}{2}b_2*(h_1+h_2)$$
and after simplification ($h_1+h_2 \neq 0$)
$$b_1 = b_2.$$
