In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums

Solution:

Given,

ar(△DRC) = ar(△DPC)

ar(△BDP) = ar(△ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(△BDP) = ar(△ARC)

⇒ ar(△BDP) – ar(△DPC) = ar(△DRC)

⇒  ar(△BDC) = ar(△ADC)

∴, ar(△BDC) and ar(△ADC) are lying in-between the same parallel lines.

∴, AB ∥ CD

ABCD is a trapezium.

Similarly,

ar(△DRC) = ar(△DPC).

∴, ar(△DRC) and ar(△DPC) are lying in-between the same parallel lines.

∴, DC ∥ PR

∴, DCPR is a trapezium.