prove a parallelogram theorem having a point inside it I have been trying to prove this theorem without success. Can anyone help me? Thank you very much! I wonder whether this requires more advance knowledge.
$ABCD$ is a parallelogram and $P$ is any point inside the parallelogram. The following relationship about the area is true: $S_1+ S_2= S_3+ S_4$

 A: Draw a line parallel to AD (equivalently BC) that passes through the point p. This line splits $S_1$ and $S_2$ into two parts, call these $S_{1L}$, $S_{1R}$, $S_{2L}$, and $S_{1R}$ where the second letter in the subscript corresponds to whether the partition of a particular triangle is to the left or right of the line passing through $p$. 
Consider the portion of the parallelogram to the left of the new line. You would see that if we draw an orthogonal line between $AB$ and the new line (whose length to be called $h_L$), then this would be orthogonal to the bases for both $S_3$, $S_{1L}$, and $S_{2L}$. Moreover, the the base for $S_3$ would be equal to the length of $AD$, i.e. $|AD|$. Thus, the area for $S_3$ would be equal to
$$
\frac{h_L \times |AB|}{2}
$$
Next consider $S_{1L}$ and $S_{2L}$. Call their base lengths $b_{1L}$ and $b_{2L}$ respectively. Note that $b_{1L}+b_{2L} = |AB|$. Their areas would sum to
$$
\frac{h_L \times b_{1L}}{2}+\frac{h_L \times b_{2L}}{2} = \frac{h_L \times (b_{1L}+b_{2L})}{2} = \frac{h_L \times |AB|}{2}
$$
which is exactly equal to the area of $S_3$. 
Similarly, the area for $S_{1R}$ and $S_{2R}$ would sum to the area of $S_4$.
Hence, we have 
$$
S_1+S_2 = S_{1L} + S_{2L}+ S_{1R} + S_{2R} = S_3+S_4.
$$
A: Draw a perpendicular to AB through point p that intersects AB and CD at points M and N respectively.
Therefore, $[ABCD] = (AB)(MN)$
$S2 = \frac{(AB)(PM)}{2}$
$S1 = \frac{(CD)(PN)}{2}$
Since $AB = CD$ we have: $S1 + S2 = \frac{(CD)(PN)}{2} + \frac{(AB)(PM)}{2} = \frac{(AB)(PM + PN)}{2} = \frac{(AB)(MN)}{2} = \frac{[ABCD]}{2}$
Since $S1 + S2 = \frac{[ABCD]}{2}$, and $S1 + S2 + S3 + S4 = [ABCD]$, it follows that $S1 + S2 = S3 + S4 = \frac{[ABCD]}{2}$
A: We use a "formula-free" cut and paste argument.
Draw the line through $P$  parallel to the horizontal sides. Draw the line through $P$ parallel to the other pair of sides.
We have divided our parallelogram into $4$ little parallelograms. Note that the lines $AP$, $BP$, $CP$, and $DP$ split their respective little parallelograms into $2$ congruent triangles.  
Use scissors to cut out the $4$ triangles that make up the region with combined area $S_1+S_2$. These can be rearranged to make up the region with combined area $S_3+S_4$. It follows that $S1+S2=S_3+S_4$.
A: What we have to prove is that $S_1+S_2=\frac{1}{2}$ $\cdot $ area of parallelogram
Area of parallelogram $= \frac{1}{2}\cdot bh$
$S_1+S_2=\frac{1}{2}[h_1\cdot b+  (h-h_1)\cdot b]     = \frac{1}{2}\cdot bh $ $\blacksquare$
