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Let chords AC and BD of a circle ω intersect at P. A smaller circle ω1 is tangent to ω at T and to segments AP and DP at E and F respectively.

(a) Prove that ray T E bisects arc ABC of ω.

(b) Let I be the incenter of triangle ACD and M be the midpoint of arc ABC of ω. Prove that MA = MI = MC.

(c) Let F* be the common point of ω1 and line EI other than E. Prove that I, F0 , D, T are concyclic.

(d) Prove that DF* is tangent to ω1. This means that F = F* , so that E, F, I are collinear

I have proved a, b and c so may assume that these are true. I need help in proving that d is true.

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  • $\begingroup$ Can you provide a drawing ? $\endgroup$ – Jean Marie Apr 3 '16 at 17:03
  • $\begingroup$ One is now provided $\endgroup$ – KingJ Apr 3 '16 at 17:28
  • $\begingroup$ You can zoom in on the picture so see the letters, ill work on making them bigger in the meantime $\endgroup$ – KingJ Apr 3 '16 at 17:37
  • $\begingroup$ @Narasimham I have uploaded another image $\endgroup$ – KingJ Apr 3 '16 at 17:49
  • $\begingroup$ Missing $B$ and $P$ $\endgroup$ – user164550 Apr 3 '16 at 18:46
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First of all, from previous parts, $\beta = \beta_1 = ... = \beta_6$.

Following part of the given solution, $\triangle MAE \sim \triangle MTA$. This leads to $\dfrac {MA}{MT} = \dfrac {EM}{AM}$. This further means $\dfrac {MI}{MT} = \dfrac {ME}{MI}$ (from part (a)). The newly formed ratio, together with the common angle $\tau$ give $\triangle MEI \sim \triangle MIT$. This further means $\lambda = \lambda_1$.

$\angle 1 = \tau + \lambda = \tau + \lambda_1 = \angle 2 = \angle 3$. Result follows from the converse of angle in alternate segment.

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  • $\begingroup$ Are you sure PF'I is the exterior angle to the correct cyclic quadrilateral? Which cyclic are you referring to? As far as I can tell, PF'I is exterior to IF'DT, not IF'TN $\endgroup$ – KingJ Apr 4 '16 at 23:45
  • $\begingroup$ @KingJ You are right. Got temporarily distracted by the complicated diagram. Will have it deleted shortly. $\endgroup$ – Mick Apr 5 '16 at 5:20
  • $\begingroup$ @KingJ Post completely re-written. See above. $\endgroup$ – Mick Apr 5 '16 at 14:18
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Here is a solution I have since found.

MCE = MAC = MTA

therefore MAE is similar to MTA

so we have ME*MT=MA^2

we also know MA=MI

so, MI^2 - ME*MT

now, MEI is similar to MTI

Angle chasing from here yeilds,

MEI = MIT = MIE +EIT = F'ID + F'DT = F'ID + F'TI + IF'T

Note also that,

MEC = AET = EF'T = IF'T

and

MEI = MEC + CEF'

Adding these together gives us

CEF' = F'ID + F'TI = F'ID + F'DI

Let DF' meet AC at P

Looking at PEF' we see that PF'E = F'DI + F'ID = PEF'

therefore PEF' is isosceles with PE = PF'

Therefore PF' is a tangent

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