Question

diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD)=ar(BOC). prove that ABCD is a trapezium.

Answer 1

Solution:

Given,

• ar(△AOD) = ar(△BOC)

To Prove,

• ABCD is a trapezium.

Proof:

• ar(△AOD) = ar(△BOC)

⇒ ar(△AOD) + ar(△AOB) = ar(△BOC)+ar(△AOB)

⇒ ar(△ADB) = ar(△ACB)

Areas of △ADB and △ACB are equal.

∴ they must lying between the same parallel lines.

∴ AB ∥ CD

∴ ABCD is a trapezium

Thanks

Answer 2

$\large\tt\red{A(∆AOD)=A(∆BOC)(given)}$

$\small\tt\purple{Adding\:A(∆AOD)+A(∆AOB)=A(∆BOC)+A(∆AOB)}$

$\small\tt\red{A(∆ABD)=A(∆ABC)}$

∆ABD and ∆ABC are on the same base AB and A(∆ABD) = A(∆ABC)

$\therefore$$\large\tt\pink{AB||DC}$

(triangles on the same base (or equal bases) and having equal areas lie between the same parallels)

$\large\tt\underline\orange{Hence,ABCD\:is\:a\:trapezium}$