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Example 1 - In fig, OA = OB and OD = OC. Show that - SAS

Example 1 - Chapter 7 Class 9 Triangles - Part 2

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Example 1 - Chapter 7 Class 9 Triangles - Part 3

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Transcript

Example 1 - Chapter 7 Class 9 - Triangles In Fig. , OA = OB and OD = OC. Show that (i) ∆ AOD ≅ ∆ BOC Given: OA = OB ...(1) and OD = OC ...(2) To prove: ΔAOD ≅ ΔBOC Proof: Line CD & AB intersect. ∠ AOD = ∠ BOC (Vertically opposite angles) ...(3) In ∆ AOD and ∆ BOC, OA = OB (From (1)) ∠ AOD = ∠ BOC (From (3)) OD = OC (From (2)) So, Δ AOD ≅ Δ BOC (SAS Congruence Rule) Example 1 In Fig. , OA = OB and OD = OC. Show that (ii) AD ∥ BC. Δ AOD ≅ Δ BOC (From 1st part) ∠ OAD = ∠ OBC (CPCT) But ∠ OAD & ∠ OBC and these form a pair of alternate angles If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel. Therefore, AD ∥ BC.

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Davneet Singh

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.