Question

# In a trapezium ABCD, E and F be the midpoints of the diagonals AC and BD respectively. Then, EF = ? (a) 12AB (b) 12CD (c) 12(AB+CD) (d) 12(AB−CD)

Solution

## (d) 12(AB−CD) Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively. Draw DE and produce it to meet AB at G. Consider DAEG and DCED ∠AEG = ∠CED (vertically opposite angles) AE = EC (E is the midpoint of AC) ∠ECD = ∠EAG (alternate angles) ΔAEG ≅ ΔCED ⇒ DE = EG → (1) And AG = CD → (2) In ΔDGB E is the midpoint of DG [From (1)] F is the midpoint of BD ∴ EF is parallel to GB ⇒ EF is parallel to AB ⇒ EF is parallel to AB and CD Also, EF = 12 GB ⇒EF = 12 (AB − AG) ⇒ EF = 12 (AB − CD) [From (2)] MathematicsSecondary School Mathematics IXStandard IX

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