# A point O inside

Construction:

Proof:

ar(Δ AOD) + ar(Δ BOC) = 1/2 (AD× BD)+ 1/2 (BC× OQ)

AD = BC (opp. Sides of rectangle)

ar(Δ AOD) + ar(Δ BOC) = 1/2 (AD× OP)+ 1/2 (AD× OQ)

= 1/2 (AB× AB)

= 1/2 ar(rect. ABCD) ….. (1)

And,

ar(Δ AOB)+ar(Δ COD) = 1/2 (AB× OS)+ 1/2 (CD× OR)

AB = CD (opp. Sides of rectangle)

ar(Δ AOB)+ar(Δ COD) = 1/2 (AB× OS)+ 1/2 (AB× OR)

= 1/2 AB× (OS+OR)

= 1/2 (AB× SR)

= 1/2 ar(rect. ABCD) …… (2)

From (1) and (2) we get,

ar(Δ AOD) + ar(Δ BOC) = ar(Δ AOB)+ar(Δ COD)

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