Median geometry

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Sedro

#1 h Apr 20, 2025, 6:03 PM • 1 Y

Y by aidan0626

In triangle $ABC$ , points $D$ , $E$ , and $F$ are the midpoints of sides $BC$ , $CA$ , and $AB$ , respectively. Prove that the area of the triangle with side lengths $AD$ , $BE$ , and $CF$ has area $\tfrac{3}{4}[ABC]$ .

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#2 h Apr 20, 2025, 7:11 PM • 1 Y

Y by Sedro

Given $\triangle ABC\ :\ A(0,2a),B(-2b,0),C(2c,0)$ , area $\ =\ 2a(b+c)$ .
Midpoints $D(c-b,0),E(c,a),F(-b,a)$ .
Distances $AD=\sqrt{(c-b)^{2}+4a^{2}},BE=\sqrt{(2c+b)^{2}+a^{2}},CF=\sqrt{(c+2b)^{2}+a^{2}}$ .

New x-axis and y-axis: $C(0,0),F(\sqrt{(c+2b)^{2}+a^{2}},0)$ .
Circle, midpoint $C$ and radius $BE$ ; second circle, midpoint $F$ and radius $AD$ .
Both circles intersect in a point $(/,\frac{3a(b+c)}{\sqrt{(c+2b)^{2}+a^{2}}})$ .
Area $\ =\ \frac{1}{2} \cdot \sqrt{(c+2b)^{2}+a^{2}} \cdot \frac{3a(b+c)}{\sqrt{(c+2b)^{2}+a^{2}}}=\frac{3}{2}a(b+c)$ .

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Sedro

#3 h Apr 20, 2025, 7:38 PM

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There is a nice synthetic solution as well.

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#4 h Apr 20, 2025, 9:47 PM • 1 Y

Y by Sedro

Because the medians divide the triangle into $6$ triangles of equal area, as well as a parallelogram's diagonals dividing itself into $4$ triangles of equal area, we have that a triangle with sides $2m_a/3, 2m_b/3, 2m_c/3$ occupies $\frac{1}{3}$ the area of $ABC.$ Thus, dilating with scale factor $3/2$ and noting that the area of a triangle is of degree $2,$ we have that a triangle with sides $m_a, m_b, m_c$ will occupy $\frac{9}{4} \cdot \frac{1}{3} = \frac{3}{4}$ the area of triangle $ABC.$

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Sedro

#5 h Apr 20, 2025, 10:01 PM

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Nice. I had a different solution: let $X$ , $Y$ , and $Z$ lie on sides $BC$ , $CA$ , and $AB$ , respectively, such that $BX:CX = CY:AY = AF:BF = 2:1$ . Notice that $XY\parallel AD$ , $YZ\parallel BE$ , and $XY\parallel AD$ . Since $AC:YC = DC:XC = 3:2$ , we have $ACD\sim YCX$ and $XY = \tfrac{2}{3}AD$ . Analogously, $YZ = \tfrac{2}{3}BE$ and $ZX = \tfrac{2}{3}CF$ . Thus, $[XYZ]$ is $\tfrac{4}{9}$ the area of the triangle with side lengths $AD$ , $BE$ , and $CF$ . It is straightforward to compute that $[XYZ] = \tfrac{1}{3}[ABC]$ , and thus the area of the triangle with side lengths $AD$ , $BE$ , and $CF$ is $\tfrac{9}{4}\cdot \tfrac{1}{3}[ABC] = \tfrac{3}{4}[ABC]$ .

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