A Lemma of Equal Areas

Bui Quang Tuan came up with a simple but surprisingly useful lemma:

Let A, B, C be three collinear points. D, E are two points on one side with respect to line that contains A, B, C and such that AD||BE and BD||CE. F is midpoint of AC. Then Area(DEF) = (Area(ABD) + Area(BCE))/2.

Observe that if F coincides with B, the claim is rather obvious, for then the three triangles DEF, ABD, and BCE are equal. Interestingly, the equality of the areas holds in a more general case.

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Proof

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Proof

In the trapezoid ABED, Area(ABD) = Area(AED) because the two triangles share a base AD and are in the same parallel lines, meaning that the altitudes to AD from B and E are equal. Similarly, in the trapezoid BCED, Area(BCE) = Area(CDE). Adding the two identities,

Area(ABD) + Area(BCE) = Area(AED) + Area(CDE).

Let h_a, h_f, h_c be the lengths of the perpendiculars to DE from A, F, and C, respectively. h_f = (h_a + h_c)/2. On the other hand, Area(AED) = DE·h_a/2 and Area(CDE) = DE·h_c/2. Therefore,

	Area(ABD) + Area(BCE)	= Area(AED) + Area(CDE)
		= DE·(h_a + h_c)/2
		= DE·h_f
		= 2·Area(DEF).

Note

If h_b is the altitude from B to DE, then - using similarities - h_b/h_a = h_c/h_b, from which Area(ADE)×Area(CDE) = Area(BDE)². As a consequence of the above, we also have Area(ABD)×Area(BCE) = Area(BDE)².

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