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Ex 8.1, 6 - Diagonal AC of a parallelogram ABCD bisects A - Ex 8.1

Ex 8.1, 6 - Chapter 8 Class 9 Quadrilaterals - Part 2

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Ex 8.1, 6 - Chapter 8 Class 9 Quadrilaterals - Part 3 Ex 8.1, 6 - Chapter 8 Class 9 Quadrilaterals - Part 4

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Ex 8.1, 6 Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that (i) it bisects ∠ C also, Given: Parallelogram ABCD where ∠ 1= ∠ 2 To prove: AC bisects ∠ C i.e. ∠ 3 = ∠ 4 Proof: Now, ∠ 1 = ∠ 2 ∠ 2 = ∠ 3 ∠1 = ∠4 Hence, we can say that ∠ 1 = ∠ 2 = ∠ 3 = ∠4 So, ∠ 3 = ∠ 4 Hence proved Ex 8.1, 6 Diagonal AC of a parallelogram ABCD bisects ∠ A Show that (ii) ABCD is a rhombus. Rhombus is a parallelogram with all sides equal ABCD is a parallelogram So, we have to prove all sides are equal In first part we proved that ∠ 1 = ∠ 2 = ∠ 3 = ∠4 Hence, ∠2 = ∠4 Now, in Δ ABC ∠ BAC = ∠ BCA Hence, BC = AB Also, AB = CD & AD = BC From (1) & (2) ⇒ AB = BC = CD = DA Hence, ABCD is a rhombus.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.