Question

A point O inside a rectangle ABCD is joined to the vertices. Prove that: ar(Δ AOD)+AR(ΔBOC)=AR(ΔAOB)+AR(ΔCOD)

Answer 1

Given: A rect∠ ABCD and a point o inside it. OA, OB, OC and OD are joined.

To prove: ar(ΔAOD)+ar(ΔBOC) = ar(ΔAOB) + ar(ΔCOD)

Construction: Draw EOF ║ AB and GOH ║AD

Proof:

        ar(ΔAOD)+ar(ΔBOC) = 1/2 (AD × EO) + 1/2(BC+OF)

                                          = 1/2(AD×EO) + 1/2(AD×OF)

                                          = 1/2 AD × (EO + OF)

                                          = 1/2 AD × EF

                                          = 1/2 AD×AB

                                         = 1/2 ar(rect∠ ABC) .............(i)

Now,

ar(ΔAOD) + ar(ΔCOD) = 1/2 (OG ×AB) + 1/2(OH×CD)

                                    = 1/2 (OG× AB) + 1/2(OH×AB) [∵ AB= CD]

                                   = 1/2 AB× GH

                                   = 1/2 AB× AD

                                   = 1/2 ar (rect∠ ABCD) .................(ii)

Hence, from equation (i)and (ii), it is proved that,

ar(Δ AOD)+ar(ΔBOC)=ar(ΔAOB)+ar(ΔCOD)

Answer 2

Answer:

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 (AD × (OP+OQ))

= 1/2 (AD× PQ)

= 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 (AB× AD)

= 1/2 ar(rect. ABCD) …… (2)

From (1) and (2) we get,

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