Ex 9.3, 8

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 ΔAEB lie on the same base BE and are between the same parallels BE and AC. ∴ ar (ΔABE) = 1/2 ar (BCNE) Similarly ∴ BCFM is a parallelogram. and ar (ΔACF) = 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)