Given, ∠COD = 120° and ∠BAC = 30°.
As ∠COD = 120°, ∠DAC = 60° (Central angle)
∠DAC + ∠BAC = 60° + 30° = 90°
∠A = 90° ⇒ ∠BCD = 90° (Opposite angles are supplementary in a cyclic quadrilateral).
Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and $ (\vec{AB}-\vec{BC})+(\vec{AD}-\vec{DC})=k \vec{FE} $, then k is equal to