Ex 9.3, 15 - Diagonals AC and BD of a quadrilateral ABCD - Triangles with same base & same parallel lines

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
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Ex 9.3, 15 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 . Given: A quadrilateral ABCD where diagonals AC & BD intersect at O & ar(AOD) = ar(BOC) To prove: ABCD is a trapezium Proof : A trapezium is a quadrilateral with one pair of opposite sides parallel Given ar(AOD) = ar(BOC) Adding ar (ODC) on both sides, Ar(AOD) + ar(ODC) = ar(BOC) + ar(ODC) ⇒ ar(ADC) = ar(BDC) Now, ΔADC and ΔBDC lie on the same base DC and are equal in area & they lie between the lines AB & DC, ⇒ AB ∥ DC In ABCD, AB ∥ DC So, one pair of opposite sides is parallel, ∴ ABCD is a trapezium Hence proved

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