The
harmonic property leads to: |BD||CD| = |AD||BC|/2

Using the intersecting chord theorem, it follows that the circle
with diameter BC and centre O, passes through A (because
of the right angle) and a point E where ADE is a straight line
with |DE| = |BC|/2, which is the radius of the circle.

/BDA = /EDO (vert opp) = /EOD (angles of isosceles
triangle EDO) = 2 /BAD (angles at
centre and circumf)
So k = 2