# Theorem 8.9

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Theorem 8.9 The line segment joining the mid-points of two sides of a triangle is parallel to the third side. Given : ABCD is a triangle where E and F are mid points of AB and AC respectively To Prove : EF ∥ BC Construction : Through C draw a line segment parallel to AB & extend EF to meet this line at D. Proof : Since EB ∥ DC with transversal ED. ∠AEF = ∠CDF In △AEF and △CDF ∠AEF = ∠CDF AF = CF ∠AFE = ∠CFD ∴ △AEF ≅ △CDF So, EA = DC But, EA = EB Hence, EB = DC Now, In EBCD, EB ∥ DC & EB = DC Thus, one pair of opposite sides is equal and parallel. Hence EBCD is a parallelogram. Since opposite sides of parallelogram are parallel. So, ED ∥ BC i.e. EF ∥ BC Hence, proved.

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.