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°

Answer 1

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°.