Side AB of parallelogram ABCD is twice as large as side AD. Point K is the midpoint of side AB. Prove that DK is an ADC angle bisector

Consider the triangle AKD.
AK = AD (by the condition of the problem), therefore this triangle is isosceles.
By the property of an isosceles triangle ∠ADK = ∠AKD
∠AKD = ∠KDC (as these are cross-angles).
It turns out that ∠ADK = ∠AKD = ∠KDC.
Therefore, DK is a bisector.

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