Given a circle with a drawn diameter and an a point K (within the circle, neither on the circumference nor on the diameter line).
Using only a straightedge draw the
perpendicular from K to the diameter .
Let the endpoints of the diameter be A and B.
1) Ray AK intersects the circle again at point C.
2) Ray BK intersects the circle again at point D.
3) Lines AD and BC intersect at point L.
4) Line KL intersects diameter AB at point F.
We need to show that /BFK = 90.
Since /ACB and /ADB are subtended by the diameter
AB they are right angles and therefore CKDL is
a cyclic quadrilateral.
/FKB = /DKL = /DCL = 90 - /DCK
= 90 - /DCA = 90 - /DBA
= 90 - /KBF
Therefore, /BFK = 90.
Posted by Bractals
on 2015-11-26 12:44:00