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If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

Rectilinear Figures

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Answer

Since, AB || CD (Opposite sides of parallelogram are parallel) and BD is transversal.

From figure,

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP. Rectilinear Figures, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

∠ABP = ∠CDQ (Alternate angles are equal)

BP = QD (As P and Q are points of trisection of the diagonal BD)

AB = CD (Opposite sides of parallelogram are equal)

Hence, △ABP ≅ △CDQ by SAS axiom.

∴ ∠APB = ∠CQD …….(By C.P.C.T.)

Multiplying both sides by -1

-∠APB = -∠CQD

Adding 180° on both sides,

⇒ 180° - ∠APB = 180° -∠CQD

⇒ ∠APQ = ∠CQP

The above angles are alternate angles and since they are equal we can say,

AP || CQ.

Hence, proved that AP || CQ.

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