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

Here, ABCD is a quadrilateral where diagonals AC and BD intersect at O.

⇒ ar(△AOD) = ar(△BOC)

Adding ar(ODC) on both sides,

⇒ ar(AOD) + ar(ODC)= ar(BOC) + ar(ODC)

⇒ ar(ADC)=ar(BDC)

Now, △ADC and △BDC lie on the same base DC and are equal in area and they lie between the lines AB and DC

⇒ AB || DC

[ Two triangles having the same base and equal areas lie between the same parallels ]

In ABCD,

⇒ AB || DC

So, one pair of opposite sides is parallel,

∴ ABCD is trapezium.

Answer 2

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Given:-

• ABCD is trapezium with AB || DC.

• Diagonals of trapezium are AC & BD.

To Prove:-

• ar(AOD) = ar(BOC)

Solution: Refer to the attachment. We have

• AB || DC

• Diagonals AC & BD intersect at point O.

Now in ∆ADB and ∆BCA

• These two triangles lie on the same base AB and are between same parallels AB & DC therefore areas of these triangles will be equal.

So, we have now

• ar (∆ADB) = ar (∆BCA)

As we can see there is a common triangle i.e ∆AOB in the center of these two ∆s.

∴ Subtracting ar (∆AOB) both the sides.

➟ ar (∆ADB) – ar (∆AOB) = ar (∆BCA) – ar (∆AOB)

➟ ar (∆AOD) = ar (∆BOC)

$\bf\green{Hence, ~proved.}$

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$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\red{\bigstar\: More \: to \; know \: :}}}}}\mid}$⠀

• A trapezium is a type of quadrilateral with 4 vertices and 4 edges.

• Area is given by 1/2 × Sum of parallel sides × Distance between them.

• Perimeter - Sum of all sides.