A $(4, - 6$ ), B $(3, - 2$ ) and C $(5, 2)$ are the vertices of a $Δ A B C$ and AD is its median. Prove that the median AD divides $Δ$ ABC into two triangles of equal areas.

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Solution

Given: AD be the median of $Δ A B C$

BD = DC

Sice D is the mid-point of BC

Using the mid point theorem to find the coordinates of D,

$D = [\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}]$

$D = [\frac{3 + 5}{2}, \frac{- 2 + 2}{2}] = [4, 0]$
Area of $Δ A B D = \frac{1}{2} | [(x_{1} (y_{2} - y_{4}) + x_{2} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{2}))] |$

$= \frac{1}{2} | [(4 (- 2 - 0) + 3 (0 - (- 6)) + 4 (- 6 - (- 2)))] |$

$= \frac{1}{2} | [- 8 + 18 - 26] |$

$= \frac{1}{2} | [- 6] | = 3$ Sq.units

Area of $Δ A D C = \frac{1}{2} | [(x_{1} (y_{4} - y_{3}) + x_{4} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{4}))] |$

$= \frac{1}{2} | [(4 (0 - 2) + 4 (2 - (- 6)) + 5 (- 6 - 0))] |$

$= \frac{1}{2} | [- 8 + 32 - 30] |$

$= \frac{1}{2} | [- 6] | = 3$ Sq.units

Area od $Δ A B D = Δ A D C$
Hence proved that the median of triangle AD divides the triangles into equal areas.

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A (4,−6 ), B (3,−2 ) and C (5,2) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.

A $(4, - 6$ ), B $(3, - 2$ ) and C $(5, 2)$ are the vertices of a $Δ A B C$ and AD is its median. Prove that the median AD divides $Δ$ ABC into two triangles of equal areas.