Question

In the given figure, D and E are two points on side BC of ∆ABC such that BD = DE = EC. Prove that ar(∆ABD) = ar(∆ADE) = ar(∆AEC).

Answer 1

We know that area of a triangle = $\frac{1}{2}$(base $\times$ height​)
Draw AL ⊥ BC.
Let h be the height of ∆ABC, AL.

∴ Height of ∆ABD ​= height of ∆ADE ​ = height of ∆AEC
The bases of ∆ABD, ∆ADE and ∆AEC are BD, DE and EC, respectively.

Given: BD = DE = EC​

∴​ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)