Question

In the adjoining figure, AD is one of the medians of a ∆ABC and P is a point on AD.
Prove that
(i) ar(∆BDP) = ar(∆CDP),
(ii) ar(∆ABP) = ar(∆ACP),

Answer 1

A median of a triangle divides it into two triangles of equal areas.
AD is a median of ∆ABC.
i.e., ar(∆ABD) = ar(∆ACD) ...(i)

(i) Now, PD is also a median of ∆PBC.
So ar(∆BDP) = ar(∆CDP) ...(ii)

(ii) Now, from (i) and (ii), we have:
ar(∆ABD) $-$ ar(∆BDP) = ar(∆ACD) $-$ ar(∆CDP)
∴ ar(∆ABP)​ = ar(∆ACP)