ABCD is a parallelogram and O is a point in its interior then show that ar(triangle AOB ) +ar(triangle COD) = ar ( triangle AOD) + ar(triangle BOC)
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ar(Δ AOB) + ar(ΔCOD) = ar(ΔAOD) + ar(ΔBOC)
Step-by-step explanation:
Given,
ABCD is a ║gm with center O where,
EF║AB, EF║DC, MN║AD, & MN║BC
In Δ AOB & ║gm ABEF,
AB = common base (since AB ║ EF)
∵ ar(Δ AOB) = 1/2 ar(║gm ABEF) ...(i)
In ΔDOC & ║gm EFCD,
DC = common base (since DC ║ EF)
∵ ar(Δ DOC) = 1/2 ar(║gm EFCD) ...(ii)
by adding (i) & (ii),
ar(Δ AOB) + ar(Δ DOC) = 1/2ar(║gm ABEF) + 1/2 ar(║gm EFCD)
= ar(Δ AOB) + ar(Δ COD) = 1/2 ar(║gm ABCD) ...(iii)
Since,
ar(Δ AOD) + ar(Δ BOC) = 1/2 ar(║gm ABCD) ...(iv)
From (iii) & (iv), we get;
ar(Δ AOB) + ar(ΔCOD) = ar(ΔAOD) + ar(ΔBOC)
Learn more: Parallelogram
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