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Question
ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: ∠EFG = 90°
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Solution
Diagonals of a kite intersect at right angles
∴ ∠MON = 90° .......(i)
In ΔBCD,
E and F are mid-points of CD and BC respectively.
Therefore, EF || DB and EF = `(1)/(2)"DB"` .......(ii)
EF || DB ⇒ MF || ON
∴ ∠MON + ∠MFN = 180°
⇒ 90° + ∠MFN = 180°
⇒ ∠MFN = 90°
⇒ ∠EFG = 90°.
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