InterviewSolution
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InterviewSolution
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In the given figure, E and F are respectively, the mid-points of non-parallel sides of a trapezium ABCD. Prove that(i) EF ║ AB (ii) EF = 1/2 (AB + DC). |
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Answer» Join BE and produce it to intersect CD produced at point P. In ∆AEB and ∆DEP, AB ║ PC and BP is transversal ⇒ ∠ABE = ∠DPE [Alternate interior angles] ∠AEB = ∠DEP [Vertically opposite angles] And AE = DE [E is mid - point of AD] ⇒ ∆AEB ≅ ∆DEP [By ASA] ⇒ BE = PE [By cpctc] And AB = DP [By cpctc] Since, the line joining the mind-points of any two sides of a triangle is parallel and half of the third side, therefore, is ∆BPC, E is mid-point of BP [As, BE = PE] and F is mid-point of BC [Given] ⇒ EF ║ PC and EF = 1/2 PC ⇒ EF ║ DC and EF = 1/2 (PD + DC) ⇒ EF ║ AB and EF = 1/2 (AB + DC) [As, DC ║ AB and PD = AB] |
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