In the adjoining figure,DE||BC.Prove that

(i) $a r (△ A C D) = a r (△ A B E)$

(ii) $a r (△ O C E) = a r (△ O B D)$

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Solution

∆DEC and ∆DEB lies on the same base and between the same parallel lines.
So, ar(∆DEC) = ar(∆DEB) ...(1)

(i) On adding ar(∆ADE) in both sides of equation (1), we get:
ar(∆DEC) + ar(∆ADE) = ar(∆DEB) + ar(∆ADE)
⇒ ar(∆ACD) = ar(∆ABE)

(ii) On subtracting ar(ODE) from both sides of equation (1), we get:
ar(∆DEC) − ar(∆ODE) = ar(∆DEB) − ar(∆ODE)
⇒ ar(∆OCE) = ar(∆OBD)

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