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(N/A) Let $\Delta ABC$ be a triangle and let $AD$ be one of its medians,where $D$ is the midpoint of $BC$.We wish to show that $\operatorname{ar}(ABD)

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(N/A) Let $\Delta ABC$ be a triangle and let $AD$ be one of its medians,where $D$ is the midpoint of $BC$.
We wish to show that $\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Since the formula for the area of a triangle involves an altitude,let us draw $AN \perp BC$.
Now,$\operatorname{ar}(ABD) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BD \times AN$.
Since $AD$ is a median,$BD = CD$.
Substituting $BD = CD$ in the expression for $\operatorname{ar}(ABD)$,we get:
$\operatorname{ar}(ABD) = \frac{1}{2} \times CD \times AN$.
Since $\frac{1}{2} \times CD \times AN$ is the formula for the area of $\Delta ACD$,we have:
$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Thus,a median of a triangle divides it into two triangles of equal areas.

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

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