Question

In figure, P is any point on median AD of ∆ABC, show that ar (∆ABP) = ar (∆ACP).

Answer 1

Here AD is median. 

Therefore ar (∆ABD) = ar (∆ACD) …(i)

Reason:

(Median divides the triangle in two parts equal in areas)

Again PD is a median of ∆BPC

ar (∆BPD) = ar (∆CPD) …(ii)

Subtracting (ii) from (i), we get

ar (∆ABD) – ar (∆BPD) = ar (∆ACD) – ar (∆CPD)

⇒ ar (∆ABP) = ar (∆ACP)