Question

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Answer 1

Given ABC is a triangle E and F are midpoints of the sides AB ,AC respectively. To prove : EF|| BC and EF = ½ BC. Construction : Draw a line CD parallel to AB ,it intersects EF at D. Proof : In a ΔAEF and ΔCDF ∠EAF = ∠FCD ( Alternative interior angles) AF = FC ( F is the midpoint) ∠AFE = ∠CFD ( vertically opp. Angles) ΔAEF ≅ ΔCDF (ASA congruence property) So that EF = DF and AE = CD ( By CPCT ) BE = AE = CD ∴ BCDE is parallelogram. ⇒ ED | | BC (Opposite sides of parallelogram are parallel) ⇒ EF | | BC ∴ EF = DF (Proved) ⇒ EF + DF = ED = BC (Opposite sides of the parallelogram are equal) ⇒ EF + EF = BC ⇒ 2 EF = BC ∴ EF = (1/2) BC.

Answer 2

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