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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.  

710815_569609_ans_88637597e2c2492daceee94efdab6fe4.jpg

Mathematics

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