The following information
involves a relationship in plane geometry attributed to Pappus.
The setup is as follows:
Consider any triangle
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
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:
is congruent to BOG because they
are corresponding angles on parallel lines.
is congruent to DAF ----> corresponding
's on parallel lines.
FA is congruent
to GB since opposite sides of parallelograms are congruent.
is congruent to DHA:
is congruent to EQC.
is congruent toQCE.
HA is congruent
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.
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
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
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.
ODQCAB is congruent to hexagon BACNYP.
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.
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