Asked by Mathslover

Circle Γ with center O has diameter AB=192. C is a point outside of Γ, such that D is the foot of the perpendicular from C to AB and D lies on the line segment OB. From C, a tangent to Γ is drawn, touching Γ at E, where the foot of the perpendicular from E to AB lies within AD. CD intersects EB at F. If CF=110, what is the length of OC?
Answered by dgewhg
145
Answered by SOmebody
wrong answer..
Answered by Alestair
156
Answered by rajeev bhatt
234
Answered by ashutosh
Make a line segment OE. Since OE=OB, angle OEB = angle OBE. This implies that angle FEC = 90 - angle OEB = 90 - angle OBE = angle DFB = angle EFC.

Thus angle FEC = angle EFC. Thus EC = CF = 110.

Now in right angle triangle CEO, CE = 110, OE = 96. By Pythagoras's theorem

OC^2 = OE^2 + EC^2 = 110^2 + 96^2 = 21316. This implies OC = 146

the correct ans is 146.....and say Thanks.....
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