Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3, 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB Given: A parallelogram ABCD where E is point on side AD produced & BE intersects CD at F To Prove: ΔABE ∼ ΔCFB. Proof: In parallelogram ABCD , opposite angles are equal, Hence, ∠A = ∠C Also, In parallelogram ABCD opposite sides are parallel, AD ∥ BC Now since AE is AD extended, AE ∥ BC and BE is the traversal ∴ ∠ AEB = ∠ CBF Now in Δ ABE & Δ CFB ∠A = ∠C ∠ AEB = ∠ CBF ∴ Δ ABE ∼ Δ CFB Hence proved

Chapter 6 Class 10 Triangles

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.