Lesson Remarkable properties of triangles into which diagonals divide a quadrilateral

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<H2>Remarkable properties of triangles into which diagonals divide a quadrilateral</H2>

<H3>Problem 1</H3>Let &nbsp;ABCD &nbsp;is a convex quadrilateral and &nbsp;AC &nbsp;and &nbsp;BD &nbsp;are its diagonals 
intersecting at point &nbsp;P &nbsp;inside the quadrilateral.
If quadrilateral is a parallelogram, &nbsp;then triangles &nbsp;APB, &nbsp;BPC, &nbsp;CPD &nbsp;and &nbsp;DPA &nbsp;all have the same area.

<B>Solution</B>

<pre>
The proof is very easy.  Consider triangles APB and BPC.

They have equal bases  |AP| = |PC|  and  common altitude h drawn from vertex B perpendicularly
to the diagonal AC. Due to the formula for the area of triangles,

{{{area[APB]}}} = {{{(1/2)*abs(AP)*h}}} = {{{(1/2)*abs(PC)*h)}}} = {{{area[BPC]}}}.

For all other pairs of triangles, the proof is similar.

At this point, the proof of the problem' statement is complete.
</pre>

<H3>Problem 2</H3>Let &nbsp;ABCD &nbsp;is an arbitrary convex quadrilateral and &nbsp;AC &nbsp;and &nbsp;BD &nbsp;are its diagonals 
intersecting at point &nbsp;P &nbsp;inside the quadrilateral.
If triangles &nbsp;APB, &nbsp;BPC, &nbsp;CPD &nbsp;and &nbsp;DPA &nbsp;all have the same area, &nbsp;then quadrilateral &nbsp;ABCD &nbsp;is a parallelogram.

<B>Solution</B>

<pre>
The proof is easy.  Consider triangles APB and BPC.

They have bases  |AP|  and  |PC|  and  common altitude h drawn from vertex B perpendicularly
to the diagonal AC.  Since they have equal areas, we can write this area equation

{{{(1/2)*abs(AP)*h}}} = {{{(1/2)*abs(PC)*h)}}}.

Canceling common factors  {{{(1/2)*h}}}  in both side, we get  |AP| = |BC|.

In other words, the intersection point P divides the diagonal AC in equal parts.

Similarly, considering triangles BPC and CPD, we prove that the intersection point P 
divides the diagonal BD in equal parts.

Next, it is a well known fact, that if the intersection point is the midpoint 
of diagonals of a quadrilateral, then the quadrilateral is a parallelogram.

At this point, the proof of the problem' statement is complete.
</pre>

Thus a quadrilateral is a parallelogram if and only if its diagonals divide
the quadrilateral in four triangles of equal areas.

This condition is a necessary and sufficient for a quadrilateral to be a parallelogram.

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