Geometry - Tangent circles

Question

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.

Answer 1

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.

Answer 2

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