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Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.3, 8 XY is a line parallel to side BC of a triangle ABC. If BE AC and CF AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF) Given: ABC Where XY BC & BE AC & CF AB To prove: ar (ABE) = ar (ACF) Proof : Let XY intersect AB & BC at M and N respectively Given that XY BC EN BC Also, it is given that BE AC BE CN In BCNE, both pair of opposite sides are parallel. BCNE is a parallelogram. Parallelogram BCNE and AEB lie on the same base BE and are between the same parallels BE and AC. ar ( ABE) = 1/2 ar (BCNE) Similarly BCFM is a parallelogram. and ar ( ACF) = 1/2 ar (BCFM) Parallelograms BCNE and BCFM are on the same base BC and between the same parallels BC and EF. ar (BCNE) = ar (BCFM) From (1) , (2) & (3) ar (ABE) = ar (ACF)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.