In $\Delta ABC$, the midpoint of $BC$ is $D$ in the given figure. Median $AD$ is drawn. $P$ is any point on $AD$.
Show that $\text{Area}(\Delta ABP) = \text{Area}(\Delta ACP)$.                                                                                                              [Page no. 143]

Given:

D is the midpoint of BC.

AD is a median.

P is any point on AD.

We know that a median divides a triangle into two triangles of equal area.

So, $\text{Area}(\Delta\text{ABD}) = \text{Area}(\Delta\text{ACD})$

Now point P lies on median AD.

Consider triangles $\Delta\text{ABP}$ and $\Delta\text{ACP}$.

Both triangles have bases BP and PC lying on the same line BC, and the same altitude from A.

Since D is midpoint of BC, the median balances the triangle equally on both sides.

Hence, $\text{Area}(\Delta\text{ABP}) = \text{Area}(\Delta\text{ACP})$

Hence, proved.