ABCD is a parallelogram.P and Q on BC trisects it.Prove that ar(APQ) =ar(DPQ)=1/6ar(ABCD)

**Given,** ABCD is a parallelogram in which points P and Q trisects the BC.

⇒ BP = PQ = QC

**To prove:**

**Construction:** Through P and Q, draw PR and QS parallel to AB and CD.

**Proof:**

Now, ΔAPD and ΔAQD lie on the same base AD and between the same parallel AD and BC.

∴ ar (ΔAPD) = ar (ΔAQD)

⇒ ar (ΔAPD) - ar (ΔAOD) = ar (ΔAQD) - ar (ΔAOD) [on subtracting ar (ΔAOD) from both sides]

⇒ ar (ΔAPO) = ar (ΔOQD) ..... (1)

⇒ ar (ΔAPO) + ar (ΔOPQ) = ar (ΔOQD) + ar (ΔOPQ) [on adding ar (ΔOPQ) on both sides]

⇒ ar (ΔAPQ) = ar (ΔDPQ) ..... (2)

Again, ΔAPQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

.... (3)

Now,

Using (2), (3) and (4), we get

[Hence proved]

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