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# You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $$\triangle$$ ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.

So, coordinates of D are ( $$\frac{3+5}{2} \ , \ \frac{-2+2}{2}$$ )
Therefore, coordinates of D = (4,0)

Now,
Area of ∆ ABD = $$\frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] |$$
= $$\frac{1}{2} \ | \ [ -8 + 18 - 16] \ |$$
= $$|-3| \ = \ 3$$ sq. units

Similarly,
Area of ∆ ACD = $$\frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ |$$
= $$\frac{1}{2} \ | \ [ -8 + 32 - 30] \ |$$
= $$|-3| \ = \ 3$$ sq. units

$$\therefore$$ Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.