Question

In the figure D, E are points on the sides AB and AC respectively of $△$ABC such that $area\left(△DBC\right)=area\left(△EBC\right)$. Prove that DE || BC.

Answer 1

Given D and E are point on sides AB and AC respectively of triangle ABC such that $Area\left(\Delta DBC\right)$ $=$ $Area\left(\Delta EBC\right)$.

Given that $Area\left(\Delta DBC\right)$ $=$ $Area\left(\Delta EBC\right)$.

Then the height of triangle DBC and Triangle EBC are the same because both triangles are the same

Then it possible that both triangle on the same base and between parallel line BC and DE

Therefore, they will lie between the same parallel lines.

$\therefore BC\parallel DE$ $\left[henceproved\right]$