EMAT 6690

Essay 3

Pappus Areas

by Kimberly Burrell, Brad Simmons, and Doug Westmoreland

Pappus of Alexandria was a Greek Mathematician.  In 320 A.D. he composed a work with the title Collection (Synagoge).  This work was very important because on several reasons.  It is the most valuable historical record of Greek Mathematics that would otherwise be unknown to us.  We are able to learn that Archimedes' discovered the 13 semiregular polyhedra, which are today known as "Archimedian solids."  He also include alternate proofs and supplementary lemmas for propositions from Euclid, Archimedes, Apollonius, and Ptolemy.  Pappus' treatise includes new discoveries and generalizations not found in early work.  The Collection contained eight books.  The first book and the beginning to book two have been lost.  In Book IV, Pappus included an elementary generalization of the Pythagorean theorem.  He also included the following problem, which has came to be known as the Pappus areas theorem.  It is not known whether or not the problem originated with Pappus, but it has been suggested that possibly it was known earlier to Heron.

Consider any triangle ABC.

 

Construct the external parallelogram on side AB.

 

Construct the external parallelogram on side AC.

 

Extend the external parallels to intersect at point P and draw segment AD.

 

Now construct the external parallelogram on side BC.

 

Now we can prove that the sum of the parallelogram areas on sides AB and AC equals the area of the parallelogram on side BC. This is what is known as Pappus Area theorem.

Proof of the Pappus Area Theorem

 

In the following sketch, one can notice that the sides of the parallelogram drawn on side BC are used to construct lines that are parallel to segment PA.  Now we can consider these triangles  PEA and JDB, and these triangles are congruent by AAS (angle-angle-side) congruency postulate.

Angle APE and Angle BJD are congruent because they are corresponding angles on a pair of parallel lines cut by a transversal, which is line GP.

Angle PEA and angle JDB are congruent also because they are corresponding angles on a pair of parallel lines cut by a transversal, which is line DP.

Side EA and side DB are congruent because they are the opposite sides of a parallelogram.

By a similar approach one can see that triangle PAF is congruent to triangle KCG.

 

Now by use of substitution, parallelogram PABJ is congruent to parallelogram EABD and parallelogram PACK is congruent to parallelogram FACG.  The sketch below indicates this relationship and shows the replacement by substitution.

 

 If we extend segment DA by constructing a line, then we will be able to construct a line through point H that is parallel to AB and a line through point I that is parallel to AC.

Angle ABC is congruent to angle LHI due to the fact that they are corresponding angles on a pair of parallel lines cut by a transversal, which is line HJ. Angle ACB is congruent to angle LIH due to the fact that they are corresponding angles on a pair of parallel lines cut by a transversal, which is line IK.  We know that BC is congruent to HI since opposite sides of a parallelogram are congruent.

From this information triangle ABC is congruent to triangle LHI by ASA (angle-side-angle) congruency postulate.  The sketch below shows this relationship.

Using substitution once again, we replace triangle ABC with triangle LHI.  This forms the hexagon BACILP that has the same area as parallelogram BCIH.

The sketch below shows this setup.

 

The corresponding angle of parallelogram JPAB and parallelogram BALH are congruent.  In this side AB is congruent to side JP and side HL.  In the original construction, the length of side BH is congruent to segment PA.  Hence, PA is congruent to AL.

Given the above, parallelogram JPAB is congruent to parallelogram BALH and parallelogram KPAC is congruent to CALI.

Thus, hexagon JPKCAB is congruent to hexagon BACILH.

Since substitution was used we can retrace the steps and conclude that the sums of the areas of parallelograms EDBA and FGCA are equal to the area of parallelogram BCIH.

 

Click here for an animated GSP sketch.

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