Question

given parallelogram ABCD, choose any point in the interior of the parallelogram. and call it X. connect point A,B,C,D to X. show the sum of areas of diagonal. nonadjacent triangles are equal. (area of triangle ABX +area triangle CDX= area triangle ADX+area triangle BCX).

Answer 1

Task. Given parallelogram ABCD, choose any point in the interior of the parallelogram and call it X. Connect point A,B,C,D to X. Show the sum of areas of diagonal nonadjacent triangles are equal.

Proof. Consider the following figure:

We should prove the sums of areas of opposite triangles are equal

$S_{ABX}+S_{CDX}=S_{ADX}+S_{BCX}.$

Since the sum of all triangles is equal to the area of parallelogram $ABCD$:

$S_{ABX}+S_{CDX}+S_{ADX}+S_{BCX}=S_{ABCD}$

it suffices to prove that one of sums is equal to the half of $S_{ABCD}$:

$S_{ADX}+S_{BCX}=\frac{1}{2}\ S_{ABCD}.$

Recall that the area of triangle with the side $a$ and the height to that side is equal to

$S=\frac{1}{2}\ ah.$

Let $XM$ be the height in the triangle $ADX$ and $XN$ be the height in the triangle $BCX$. Then

$S_{ADX}=\frac{1}{2}AD*XM,\qquad S_{BCX}=\frac{1}{2}BC*XN.$

Since $ABCD$ is a parallelogram, $AD=BC$. Moreover, the sides $AD$ and $BC$ are parallel, so $MN$ is perpendicular to $AD$ and $BC$, and so $MN$ is the height in the parallelogram $ABCD$. In particular, $X\in MN$, and thus $MN=XN+XM$. Therefore

$S_{ADX}+S_{BCX}$ $=\frac{1}{2}AD*XM+\frac{1}{2}BC*XN=\frac{1}{2}AD*XM+\frac{1}{2}AD*XN=$

$=\frac{1}{2}AD(XM+XN)=\frac{1}{2}AD*MN.$

On the other hand, it is known that the area of the parallelogram is

$S_{ABCD}=AD*MN,$

whence

$S_{ADX}+S_{BCX}=\frac{1}{2}AD*MN=\frac{1}{2}S_{ABCD}.$

Therefore

$S_{ABX}+S_{CDX}=S_{ABCD}-(S_{ADX}+S_{BCX})=S_{ABCD}-\frac{1}{2}S_{ABCD}=\frac{1}{2}S_{ABCD}=S_{ADX}+S_{BCX}.$