Lesson Remarkable properties of triangles into which diagonals divide a quadrilateral

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Remarkable properties of triangles into which diagonals divide a quadrilateral

Problem 1

Let ABCD is a convex quadrilateral and AC and BD are its diagonals
intersecting at point P inside the quadrilateral.
If quadrilateral is a parallelogram, then triangles APB, BPC, CPD and DPA all have the same area.

Solution

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,

     =  =  = .


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


At this point, the proof of the problem' statement is complete.

Problem 2

Let ABCD is an arbitrary convex quadrilateral and AC and BD are its diagonals
intersecting at point P inside the quadrilateral.
If triangles APB, BPC, CPD and DPA all have the same area, then quadrilateral ABCD is a parallelogram.

Solution

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 

     = .


Canceling common factors    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.

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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