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

Ex 9.3, 15 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2


Question 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.