In ∆ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (∆ABP) Class 9

In ∆ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (∆ABP) Class 9

Question 1.

In ∆ABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP). Class 9

Solution:

Given: AD is median of ∆ABC.

To prove: A(∆ABP) = A(∆ACP)

Proof:

In ∆ABC, AD is median of ∆ABC ... [Given]

∴ A(∆ABD) = A(∆ACD) ....(i) [Median of a triangle divides it into two triangles of equal area.

Also, In ∆BPC, PD is median of ∆BPC,

∴ A(∆PDB) = A(∆PDC) ....(ii) [Median of a triangle divides it into two triangles of equal area]

Subtracting equation (ii) from (i), we get

A(∆ABD) - A(∆PDB) = A(∆ACD) - A(∆PDC)

∴ A(∆ABP) = A(∆ACP)