Question

In the adjoining figure, D and E are points on the sides AB and AC of ∆ABC such that ar(∆BCE) = ar(∆BCD). Show that DE || BC.

Answer 1

∆BEC and ​∆BCD lies on the same base BC and have equal areas.
Darw EM as the altitude of ∆BEC and DL as the altitude of ∆BCD.
ar(∆BCE) = $\frac{1}{2}$ ⨯ base ⨯ height = $\frac{1}{2}$ ⨯ BC ⨯EM
ar(∆BCD) = $\frac{1}{2}$ ⨯ base ⨯ height = $\frac{1}{2}$ ⨯ BC ⨯DL
However, ar(∆BCE) = ar(∆BCD)
⇒ $\frac{1}{2}$ ⨯ BC ⨯EM = $\frac{1}{2}$ ⨯ BC ⨯DL
⇒ ​EM = DL
i.e., the corresponding altitudes of both the triangles are equal.
∴ DE || BC