Question

diagonal CD is bisected by diagonals AB proove that ar. of ∆ABC=ar. of ∆ADB

Answer 1

Hii...

Here is your answer....

In ∆ ACD

O is the mid point of CD

So AO is the median of ∆ ACD

Median divide a ∆ into 2 equal areas

So, ar (∆ADO ) = ar (∆ACO) ..........(1)

Similarly ar(∆BDO) = ar(∆BCO)........(2)

Adding (1 ) & (2)

ar(∆ADO) + ar(∆BDO) = ar(∆ACO) + ar(CBCO)

Now we will get

Ar(∆ADB) = ar(∆ABC)......Hence proved

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