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.
Hi, it looks like you're using AdBlock :(
Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.
Please login to view more pages. It's free :)
Teachoo gives you a better experience when you're logged in. Please login :)
Solve all your doubts with Teachoo Black!
Teachoo answers all your questions if you are a Black user!