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B D is one of the diagonals of a quad...

`B D` is one of the diagonals of a quadrilateral `A B C Ddot\ \ A M\ a n d\ C N` are the perpendiculars from `A\ a n d\ C` , respectively, on `B Ddot` Show That `a r\ (q u a ddotA B C D)=1/2B Ddot(A M+C N)`

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Given: `BD` is one of the diagonals of a quadrilateral `ABCD`. `AM` and `CN` are the perpendiculars from `A` and `C` respectively.
To prove: `ar(`quad.`ABCD)=1/2BDxx(AM+CN)`
Proof: `ar(`quad.`ABCD)=ar(/_\ABD)+ar(/_\BCD)`
`ar(`quad.`ABCD)=1/2(BDxxAM)+1/2(BDxxCN)`
`therefore ar(`quad.`ABCD)=1/2BDxx(AM+CN)`
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