Question

# Show that the median of a triangle divides it into two triangles of equal areas.

Solution

## Let ABC be a triangle and Let AD be one of its medians.In $$\triangle ABD$$ and $$\triangle ADC$$ the vertex is common and these bases BD and DC are equal.Draw $$AE \perp BC.$$Now $$area (\triangle ABD) = \dfrac{1}{2} \times base \times altitude\ of \triangle ADB$$$$= \dfrac{1}{2} \times BD \times AE$$$$= \dfrac{1}{2} \times DC \times AE (\because BD = DC)$$but DC and AE is the base and altitude of $$\triangle ACD$$$$= \dfrac{1}{2} \times$$ base DC $$\times$$ altitude of $$\triangle ACD$$$$= area \triangle ACD$$$$\Rightarrow area (\triangle ABD) = area (\triangle ACD)$$Hence the median of a triangle divides it into two triangles of equal areas.  Mathematics

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