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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(ABCD)    


Solution

(d) 12(ABCD)

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)]


Mathematics
Secondary School Mathematics IX
Standard IX

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