Pappus Area

by

Vicki Tarleton and Michelle Jones

The following information involves a relationship in plane geometry attributed to Pappus.

The setup is as follows:

Consider any triangle ABC.

Construct external parallelograms on sides AB and AC.

Extend the external parallels to intersect at point D and construct segments AD.

Use the direction and length of segment AD to construct a parallelogram on side BC.

The sum of the areas of parallelograms FABG and HACE are equivalent to the area of parallelogram BCNP.

The GSP construction below shows this relationship and the proof follows.

In the figure shown below, the sides of parallelogram BCNP have been used to construct lines, which are also parallel to segment DA. Now we can consdier the 's OGB and DFA. These triangles are congruent by the AAS (angle-angle-side) congruency postulate:

ADF is congruent to BOG because they are corresponding angles on parallel lines.

OBG is congruent to DAF ----> corresponding 's on parallel lines.

FA is congruent to GB since opposite sides of parallelograms are congruent.

Likewise, QEC is congruent to DHA:

HDA is congruent to EQC.

DAH is congruent toQCE.

HA is congruent to CE.

Therefore by substitution, parallelogram DABO is congruent to parallelogram FABG and parallelogram DACQ is congruent to paralleloram HACE. The figure below indicates this relationship and shows the replacement by substitution.

Let's next consider extending segment DA by contructing a line. Then we will construct a line through point P parallel to AB and a line through point N parallel to AC.

Now, ABC is congruent to YPN and ABC is congruent to YNP, since these are corresponding angles on parallel lines. We know that segment BC is congruent to segment PN since opposite sides of a parallelogram are congruent.

Therefore BAC is congruent to PYN. The figure below shows this relationship.

Once again, by substituting PYN for BAC, the hexagon BACNYP formed has the same area as parallelogram BCNP.

The figure below shows the setup.

Now, the corresponding angles of parallelogram ODAB and parallelogram BAYP are congruent. Also, side AB is congruent to side OD and side PY. In the original construction, the length of side BP is congruent to segment DA. Therefore DA is congruent to AY.

Given the above, parallelogram ODAB is congruent to parallelogram BAYP and parallelogram DQCA is congruent to parallelogram ACNY.

Therefore hexagon ODQCAB is congruent to hexagon BACNYP.

Since substitution was used we can retrace the steps and conclude that the sums of the areas of parallelograms FABG and HACE are equal to the area of parallelogram BCNP.