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A Relation between Altitudes of Four Triangles

​
D
M
DN
≈
4.11
PF
≈
5.30
MQ
≈
3.52
AH
≈
12.93
DN
+
PF
+
MQ
≈
12.93
Let ABC be a triangle and DEF be parallel to AC and FGM be parallel to AB. Let AH, DN, FP, and MQ be altitudes in triangles ABC, DBE, FEG, and MGC respectively. Then
AH=DN±PF+MQ
with the plus sign when F is outside of ABC and minus when F is inside.
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