Chapter 6 Class 10 Triangles (Term 1)

Serial order wise

Last updated at Aug. 4, 2021 by Teachoo

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 π΄πΈ/πΈπ· = π΅πΉ/πΉπΆ 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: π΄πΈ/πΈπ·=π΅πΉ/πΉπΆ Proof: Given AB II DC & EF II AD So, EF II DC Joining A & C Let AC intersect EF at point G Now in β π΄π·πΆ EG II DC So, π΄πΈ/πΈπ·=π΄πΊ/πΊπΆ Similarly , in β πΆπ΄π΅ AB II GF So, π΄πΊ/πΊπΆ=π΅πΉ/πΉπΆ From (1) and (2) π΄πΈ/πΈπ·=π΄πΊ/πΊπΆ=π΅πΉ/πΉπΆ π΄πΈ/πΈπ·=π΅πΉ/πΉπΆ Hence Proved