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  1. Chapter 8 Class 9 Quadrilaterals
  2. Serial order wise
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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.

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