[asy][asy] unitsize(3cm); pointpen=pathpen=black; pair B=D("B",dir(0),E), A=D("A",-B,W), C=D("C",dir(70),NE), D=D("D",foot(C,A,B)), I=D("I",incenter(A,B,C),NW), X=D("X",foot(I,A,B),SW), E=D("E",rotate(-90,X)*I,SE), F=D("F",IP(D(I--E,dashed),D(C--D)),SSW), J=D(E+F-D), G=D("G",IP(L(A,F,-0.1,2),unitcircle),NE); DPA(A--B--C--cycle^^E--J--F^^I--X^^CP(J,E)^^arc(origin,1,0,180)); D(A--G,dotted); [/asy][/asy] (a) By homothety at $G$, we have $k\to (ABC)$ and $F\to A$ (since $CD$ is parallel to the tangent at $A$), so the result follows. (b) Let $I$ be the incenter of $\triangle ABC$ and $X$ the foot from $I$ to $AB$. Then $E,F,I$ are collinear (well-known), so $XE=IX=r$. Let $R$ be the radius of $k$. Then (with directed lengths), $$R=XE-XD=r-(XB-DB)=\frac{\Delta}{s}-(s-b-a\cos B)=\frac{ab}{s}-\left(s-b-\frac{a^2}{c}\right).$$

Another problem to that configuration that appeared in the 3rd round of this year's Germany's Olympiad. It was the hardest problem on the contest. (Round 3 isn't very difficult) Prove $AC=AE$ in the above configuration.

Kezer wrote: Another problem to that configuration that appeared in the 3rd round of this year's Germany's Olympiad. It was the hardest problem on the contest. (Round 3 isn't very difficult) Prove $AC=AE$ in the above configuration. We use the same diagram in jlammy's post. Since $DFGB$ is cyclic, we have $AF \times AG = AD \times AB$. From power of a point, $AF \times AG = AE^2$. So, $AE^2 = AD \times AB = AC^2 \implies AE = AC$ as desired.

Nice problem,..., here is the sketch of my proof: (a) ** Let I be the center of k, M the midpoint of AB and N the midpoint of arc AB which doesn't contain C. Clearly M, I, G are collinear, and triangles GIE and MGN are similar which implies N, E, G are collinear. ** FIED is a square. ** Note that angFGE=angFGN=angAGN=45º , and conclude that A,F, G are collinear. (b) ** After inversion with center at A and radius AE point C remains fixed then AE=AC ** r=AE-AD=AC-AD=AC-(AC^2/AB)

We can solve part a) like this:Let K be the center of circle EFG and L midpoint of AB.Then triangles ALG and FKG are isosceles triangles,they also have common angle GKF=GLA (because L,K,G are colinear and KF is paralel to BA) and share same line LG,so AG and FG need to be on same line too.That follows A-F-G.

a) just take the reflection of C over AB that will be on the circumcircle of ABC, notice that A will be the midpoint of TC, so G-F-A. b)It is well known that $x=\sqrt{R^2-Rr}$, where $x$ is the radius of the circumcircle of triangle $\triangle{ADT}$, $R$ is the radius of the circumcircle of $\triangle{ABC}$, $r$ is the radius of the circumcircle of $k$. And since we can find easily the sides of $\triangle{ADT}$ is terms of $\triangle{ABC}$ the result follows.

I'll provide a solution that's a little more motivated and doesn't need some special facts. Notice that from $\triangle ACD \sim \triangle ABC$ and $\triangle AFD \sim \triangle ABG$, the problem is essentially equivalent to proving that $AC=AE$ (because this would mean $AE^2=AC^2=AB \cdot AD = AF \cdot AG'$, where $G' = AF \cap (ABC)$, implying that $G'$ is on (FEG), and thus $G=G'$. Construct $E'$ on segment $AB$ such that $AE'=AC$ and let $F'$ be on segment $CD$ such that $F'DE'$ is an isosceles right triangle (see why?) and let $O$ be the point such that $F'OE'D$ is a square. Let $\omega$ be the circle with center $O$ and radius $OF'=OE'$. Finally, let $G'$ be the intersection of $AF$ and $\omega$. We will prove that $\omega$ is tangent to $(ABC)$ at $G'$. We have that $AC^2=AE'^2 = AF' \cdot AG'$, so $\triangle ACF' \sim \triangle AG'C$. So $\angle AG'C = \angle ACF' = \angle ACD = \angle ABC$, so that $G' \in (ABC)$. Now let $l$ be the tangent to $(ABC)$ at $G'$. The angle that $AG'$ makes with $l$ is equal to $\angle ABG' = \angle AF'D = \angle AF'E' - 45^{\circ} = \angle AE'G' - 45^{\circ} = \angle F'E'G'$, so it follows that $l$ is tangent to $\omega$ as well. Thus, $E=E'$, $F=F'$, and $G=G'$ and we are done!