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# A(7, −3), B(5, 3), C(3, −1) are the vertices of a $∆ABC$ and AD is its median. Prove that the median AD divides $∆ABC$ into two triangles of equal areas.

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Solution

## The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1). $\mathrm{Coordinates}\mathrm{of}D=\left(\frac{5+3}{2},\frac{3-1}{2}\right)=\left(4,1\right)$ For the area of the triangle ADC, let $A\left({x}_{1},{y}_{1}\right)=A\left(7,-3\right),D\left({x}_{2},{y}_{2}\right)=D\left(4,1\right)\mathrm{and}C\left({x}_{3},{y}_{3}\right)=C\left(3,-1\right)$. Then $\mathrm{Area}\mathrm{of}∆ADC=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[7\left(1+1\right)+4\left(-1+3\right)+3\left(-3-1\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[14+8-12\right]=5\mathrm{sq}.\mathrm{unit}$ Now, for the area of triangle ABD, let $A\left({x}_{1},{y}_{1}\right)=A\left(7,-3\right),B\left({x}_{2},{y}_{2}\right)=B\left(5,3\right)\mathrm{and}D\left({x}_{3},{y}_{3}\right)=D\left(4,1\right)$. Then $\mathrm{Area}\mathrm{of}∆ABD=\frac{1}{2}\left[{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[7\left(3-1\right)+5\left(1+3\right)+4\left(-3-3\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[14+20-24\right]=5\mathrm{sq}.\mathrm{unit}$ Thus, $\mathrm{Area}\left(∆ADC\right)=\mathrm{Area}\left(∆ABD\right)=5\mathrm{sq}.\mathrm{units}$. Hence, AD divides $∆ABC$ into two triangles of equal areas.

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