Ex 9.3, 3
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Given: A parallelogram ABCD
With diagonals AC & BD
To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)
Proof : ABCD is a parallelogram
Diagonals of a parallelogram bisect each other
∴ O is the mid-point of BD, i.e., OB = OD
& O is the mid-point of AC, i.e., OA = OC
In ∆ ABC,
Since OA = OC
∴ BO is the median of ∆ ABC
⇒ ar(∆ AOB) = ar(∆ BOC)
In ∆ ADC,
Since OA = OC
∴ DO is the median of ∆ ADC
⇒ ar(∆ AOD) = ar(∆ COD)
Similarly,
In ∆ABD,
Since OB = OD
∴ AO is the median of ∆ ABD
⇒ ar(∆ AOB) = ar(∆ AOD)
From (3) , (4) & (5)
ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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