# 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

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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, S_{1}, S_{2}, S_{3} 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

S

_{1}× S_{2}= S_{3}× S_{3}.Triangle ABC, being split into four, we have

S | = S_{1} + S_{2} + 2S_{3} | |

= S_{1} + S_{2} + 2√[S_{1}·S_{2}] | ||

= (√S_{1} + √S_{2})². |

On the other hand,

4S_{3} | = (√S_{1} + √S_{2})² - (√S_{1} - √S_{2})² | |

= S - (√S_{1} - √S_{2})². |

Hence S_{3} is maximum iff S_{1} = S_{2}, i.e. only when X is the midpoint of BC and XB'C' is medial triangle.

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Copyright © 1996-2018 Alexander Bogomolny