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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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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