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

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

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

Answer»

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)