Lines AK and DM intersect at point R. Lines BK and CM
intersect at point Q.
It is easy to show that 1) point P is the midpoint of line
segment QR, 2) point Q lies on side AD, 3) point L lies
on line segment QR, 4) line segment QR is parallel to
side AB, and 5) |KM| = (2/3)|AD|.
[AKL] + [DLM] = [RAD] - [RKM] - [LKM] - [LAD]
= 6h|AD| - 4h|KM| - h|KM| - h|AD|
= 5h|AD| - 5h|KM|
= 5h|AD| - 3h|KM| - 2h|KM|
= 5h|AD| - 3h(2/3)|AD| - 2h|KM|
= 3h|AD| - 2h|KM|
= 3h|BC| - 2h|KM|
= [QBC] - [QKM]
= [BCMK]
Note: The altitudes of the six triangles are multiples
of 2h ( where the value of h is not needed for
the proof ).
QED
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