In the adjoining figure, DE || BC. Prove that
(i) ar(∆ACD) = ar(∆ABE),
(ii) ar(∆OCE) = ar(∆OBD),

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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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In the adjoining figure, DE || BC. Prove that (i) ar(∆ACD) = ar(∆ABE), (ii) ar(∆OCE) = ar(∆OBD),

In the adjoining figure, DE || BC. Prove that
(i) ar(∆ACD) = ar(∆ABE),
(ii) ar(∆OCE) = ar(∆OBD),