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

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Copyright © 1996-2018 Alexander Bogomolny

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
  = 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.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
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