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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.