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The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.

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Since AB ∥ CD and AD ∥ BC, alternate angles are equal. Because AC bisects ∠A, adjacent angles match, proving AC bisects ∠C. In △ABC, equal base angles imply AB = BC, making all sides equal.

Cbse released Class 9 Ganita Manjari Part 2 book
Class 9 Chapter 12 Quadrilaterals (Ganita Manjari 2) Solutions

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1 Answer

  1. Let diagonal AC bisect ∠A, so ∠DAC = ∠BAC. By alternate interior angles, ∠BAC = ∠DCA and ∠DAC = ∠BCA. Hence, ∠DCA = ∠BCA, proving AC bisects ∠C. Furthermore, in △ABC, ∠BAC = ∠BCA implies AB = BC. Since adjacent sides of this parallelogram are equal, all four sides are equal, confirming ABCD is a rhombus.

     

    For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/

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