1. Chapter 8 Class 9 Quadrilaterals (Term 2)
  2. Serial order wise
  3. Theorems

Theorem 8.1 - Chapter 8 Class 9 Quadrilaterals (Term 2)

Last updated at Aug. 25, 2021 by

Theorem 8.1 - Class 9 - Diagonal of parallelogram divides.jpg

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Theorem 8.1 - Chapter 8 Class 9 Quadrilaterals - Part 2

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  1. Chapter 8 Class 9 Quadrilaterals (Term 2)
  2. Serial order wise

    3. Theorems

    Previous topics →

Transcript

Theorem 8.1: A diagonal of a parallelogram divides it into two congruent triangles Given: A parallelogram ABCD with AC as its diagonal To prove: ΔABC ≅ ΔADC Proof: Opposite sides of parallelogram is parallel So, AB ∥ DC and AD‖∥ BC In Δ ABC & Δ ADC ∠ BAC = ∠ DCA AC = AC ∠ DAC = ∠ BCA ∴ ΔABC ≅ ΔADC Hence proved

Chapter 8 Class 9 Quadrilaterals (Term 2)
Serial order wise

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.