A Maximum Area Property of the Medial Triangle

Let X be a point on side BC of ΔABC. Points B' on AC and C' on AB are such that XB'||AB and XC'||AC. Then the area of ΔXB'C' achieves its maximum for X being the midpoint of BC.

To remind, the medial triangle in ΔABC is formed by the midpoints of the sides.

Proof

Let X be a point on side BC of ΔABC. Points B' on AC and C' on AB are such that XB'||AB and XC'||AC. Then the area of ΔXB'C' achieves its maximum for X being the midpoint of BC.

Proof

Let, for convenience, S, S₁, S₂, S₃ denote the areas of triangles ABC, XBC', XB'C, XB'C'. (Note that triangles XB'C' and AB'C' are congruent and thus have equal areas.)

From a note to another Bui Quang Tuan's lemma

S₁ × S₂ = S₃ × S₃.

Triangle ABC, being split into four, we have

	S	= S₁ + S₂ + 2S₃
		= S₁ + S₂ + 2√[S₁·S₂]
		= (√S₁ + √S₂)².

On the other hand,

	4S₃	= (√S₁ + √S₂)² - (√S₁ - √S₂)²
		= S - (√S₁ - √S₂)².

Hence S₃ is maximum iff S₁ = S₂, i.e. only when X is the midpoint of BC and XB'C' is medial triangle.

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