Last updated at Aug. 25, 2021 by

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 10 years. He provides courses for Maths and Science at Teachoo.