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Example 2 - ABCD is a trapezium with AB || DC. E & F are - Theorem 6.1

Example 2 - Chapter 6 Class 10 Triangles - Part 2
Example 2 - Chapter 6 Class 10 Triangles - Part 3

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Transcript

Example 2 ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig. 6.14). Show that 𝐴𝐸/𝐸𝐷 = 𝐡𝐹/𝐹𝐢 Given: ABCD is a trapezium where AB II DC E and F are points non parallel sides AD and BC such that EF II AB To Prove: 𝐴𝐸/𝐸𝐷=𝐡𝐹/𝐹𝐢 Proof: Given AB II DC & EF II AD So, EF II DC Joining A & C Let AC intersect EF at point G Now in βˆ† 𝐴𝐷𝐢 EG II DC So, 𝐴𝐸/𝐸𝐷=𝐴𝐺/𝐺𝐢 Similarly , in βˆ† 𝐢𝐴𝐡 AB II GF So, 𝐴𝐺/𝐺𝐢=𝐡𝐹/𝐹𝐢 From (1) and (2) 𝐴𝐸/𝐸𝐷=𝐴𝐺/𝐺𝐢=𝐡𝐹/𝐹𝐢 𝐴𝐸/𝐸𝐷=𝐡𝐹/𝐹𝐢 Hence Proved

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.