Geometry - Tangent circles 
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.
 A: 
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.
A: 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 
