Question

△𝐀𝐁𝐂 and parallelogram 𝐀𝐁𝐄𝐅 are on the same base, 𝐀𝐁 as in between the same parallels 𝐀𝐁 and 𝐄𝐅. Prove that 𝐚𝐫(△𝐀𝐁𝐂) = $\frac{1}{2}$ 𝐚𝐫(∥𝐠𝐦 𝐀𝐁𝐄𝐅).

Answer 1

We know that,
Parallelograms on the same base and between the same parallels are equal in area.
ar(∥gm ABEF) = ar(∥gm ABHC) .....(1)

Consider ∥gm ABHC.
BC is the diagonal. We know that, the diagonal always divides the parallelogram into 2 congruent triangles.

And we know that if two triangles are congruent, then they will have the same area and perimeter.
ar(△ABC) = ar(△BCH) .....(2)
ar(∥gm ABHC) = ar(△ABC) + ar(△BCH)
ar(∥gm ABHC) = 2ar(△ABC) ...(from (2))
ar(△ABC) = $\frac{1}{2}$ar(∥gm ABHC) .....(3)

From (1) and (3),
ar(△ABC) = $\frac{1}{2}$ar(∥gm ABEF)