In a triangle ABC , ∠BAC = 90^o. Point D lies on the side BC, and satisfies
∠BDA = k*∠BAD, where k is a real constant.

Find the value of k, given that length of AD is the harmonic mean between the respective lengths of BD and CD.

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

Posted by Harry on 2015-06-16 10:16:51