Let A , B , C be 3 points on a circle $\omega$ . let the tangents at A and at B meet at D . Let DC meets the circle again at E . Then prove that the line AE bisects BD iff AB = BC .
I have done at lot of work on it and as one might see that if I prove that right angles or if F'D = F'B then only the problem is solved but can't get it .Image of the same is pasted here
Let AE meets BD at F.
Let m arcAEB = m arcBC. (1) (m arcXYZ is for measure
of arc XYZ)
Let compare △DEF with △ADF.
∡EFD is common (2)
m∡ADB = (m arcACB - m arcAEB) / 2 (3)
Taking into account (1), we have
m∡ADB = (m arcACB - m arcBC) / 2 = (m arcAC) / 2 (4)
m∡DEF = m∡AEC = (m arcAC) / 2 (5)
From (4) and (5) we have
m∡ADB = m∡DEF (6) (m∡XYZ is for measure
of angle XYZ)
Now from (2) and (6) we have that
△DEF and △ADF are similar so we can write:
DE / AD = EF / DF = FD / FA (7)
From (7) we have
DF² = EF * FA (8)
But tangent secant segment theorem gives
BF² = EF * FA (9)
From (8) and (9) we have
DF = FB, qed.
