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Theorem 8.10 - Line drawn through mid-point of one side of triangle.jpg

2 Theorem 8.10 - Since Opposite sides of Parallelogram are equal Eb = DC.jpg
3 Theorem 8.10 - Hence F is mid point of AC Hence Proved.jpg

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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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