# Theorem 8.10

Last updated at Nov. 23, 2017 by Teachoo

Last updated at Nov. 23, 2017 by Teachoo

Transcript

Theorem 8.10 The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. Given : Δ ABC where E is mid point of AB , F is some point on AC & EF ∥ BC To Prove : F is a mid point of AC. Construction : Through C draw CM ∥ AB Extend EF and let it cut CM at D. Proof: In quadrilateral EBCD ED ∥ BC & EB ∥ CD ∴ Since both pairs of opposite sides are parallel. EBCD is a Parallelogram Since opposite sides of parallelogram are equal. EB = DC But, EB = EA Hence EA = DC Also, EB ∥ DC with transversal ED ∴ ∠AEF = ∠CDF In △AEF and △CDF ∠AEF = ∠CDF ∠AFE = ∠CFD AE = CD ∴ △AEF ≅ △CDF So, AF = CD Hence, F is a mid point of AC Hence proved.

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .