Lines Around a Circle

Draw the line tangent to $\Gamma$ at $C$ . Let the point of intersection between the tangent to $\Gamma$ at $C$ and the tangent to $\Gamma$ at $B$ be $G$ . We have that $\angle CGB=\angle ADB$ since $AB=BC$ . Also, $\angle ACD= \angle FAD$ by the alternate secant theorem. Thus, triangle $ADF$ is similar to triangle $DGC$ .

$AD=DB=BG=GC$ since $AB=BC$ . By similar triangles $ADF$ and $DGC$ , we get that $\frac {DG}{GC}=\frac {AD}{DF}=2$ . Thus, $FD=120/2=60$ and $FB=DB-FD=120-60=60$ .

We will prove that F is the midpoint of DB by showing that $FD^2=FB^2=FE.FA$

Since FB is tangent to the circle $\Gamma, FB^2=FE.FA$ (Power of point F with respect to circle $\Gamma$ ) (1)

Since DA and DB are both tangent to circle $\Gamma, DA = DB$ $\rightarrow \angle ADB = 180^\circ - 2 \angle DAB$ $= 180^\circ - 2\angle ACB$ (Alternate segment theorem)

$= \angle ABC$ (Triangle ABC is isosceles as given in the question)

$= \angle AEC$ (Angles substituted by the same arc) $= \angle DEF$

Therefore, Triangle FDA $\sim$ Triangle FED $\rightarrow \frac{FD}{FE} = \frac{FA}{FD}$ $\rightarrow FD^2 = FE.FA$ (2)

From (1) and (2) $\rightarrow$ F is the midpoint of DB

Hence $FB = \frac{1}{2}DB = \frac{1}{2}DA = 60$

Let $\angle BCA=x$ . Since $AB=BC$ , $\angle CAB=x$ . Since $BD$ is a tangent to the circle at $B$ , $\angle DBA=\angle BCA=x$ . Therefore, $\angle DBA= \angle CAB$ , $DB$ is parallel to $AC$ .

Let $\angle DCA=y$ . Since $AD$ is a tangent to the circle at $A$ , $\angle FAD=y$ . Since $BD$ and $AC$ are parallel, $\angle FDE=\angle BDC=y$ . $\angle FDE=\angle FAD$ , $\angle EFD=\angle DFA$ , triangle $EFD$ is similar to triangle $DFA$ . Consequently, $\frac {DF}{FA}=\frac {EF}{DF}$ , $DF^2=EF\times FA$ .

By Power of Point Theorem, $BF^2=EF*FA$ . $DF^2=BF^2$ . Since both $DF$ and $BF$ are positive, $DF=BF$ . Since $BF+DF=BD=AD=120$ , $BF=DF=60$ .

[Latex Edits - Calvin]

Denote $AB = BC = a$ , $BD = c = 120$ and $AC = b$ . Since angles $\angle DAB = \angle DBA = \angle BCA = \angle BAC$ , triangles ADB and ABC are similar. Thus $\frac c a = \frac a b$ .

Pick point P such that PACB is a parallelogram, so $AP = BC = a$ . Now $\frac {BD}{BC} = \frac c a = \frac a b = \frac {AP}{AC}$ ; together with $\angle PAC = \angle DBC$ , we thus see that triangles PAC and DBC are similar.

This gives the critical fact: $\angle PCA = \angle DCB$ .

Next, since AEBC is cyclic, $\angle FAB = \angle EAB = \angle ECB = \angle DCB$ , which by the above is equal to $\angle PCA$ . Let Q be the intersection of PC and AB. From the equality of angles $\angle QCA = \angle FAB$ and $\angle QAC = \angle FBA$ we see that triangles QCA and FAB are similar, so:

$\frac{FB}{QA} = \frac{AB}{CA} \implies \frac{FB}{a/2} = \frac{a}{b} = \frac{c}{a} \implies FB = \frac c 2 = 60$ .

Angle DAB = angle DBA = angle BCA = angle BAC.

Since angle DBA = angle BAC, lines BD and AC are parallel and angle FDE = angle DCA.

Angle FAD = angle DCA because they all corresponds to the same arc AE. So angle FAD = angle FDE.

So, triangle DEF is similar to triangle ADF, we can get DF^2 = AF * EF.

Meanwhile, triangle FBE is also similar to triangle FAB since they share angle BFE and angle FBE = angle FAB because they corresponds to arc BE.

So we can get BF^2 = AF * EF.

Since BF^2 = AF * EF = DF^2, BF = DF so BF = 120/2 = 60

We'll prove that $BF = DF$ .

Let $M$ and $r$ be the centre and the radius of the circle $\Gamma$ , respectively. Let $Z \in BD$ such that $\angle ABZ$ is obtuse.

First, we prove $\angle CEB = \angle DBA$ . It's clear that $\angle DBA + \angle ABM = \frac{\pi}{2}$ . Since $AB = BC$ , triangles $MBA$ and $MBC$ are congruent by SSS Property, $MA = MB = MC = r$ Thus, $\angle ABM = \angle ACM$ . Hence, $\angle CBZ = \angle DAB = \frac{\pi}{2} - \angle ABM$ .

By Alternate Segment Theorem, $\angle CBZ = \angle CEB$ , since points $C$ , $E$ , $B$ lie on circle $\Gamma$ and $B$ is the point of tangency.

Let $T$ be the intersection of lines $AB$ and $CD$ . Then, looking at triangles $TEB$ and $TBD$ , it's clear that $\angle ABE = \angle TBE = \angle EDB$ . Again, by Alternate Segment Theorem, $\angle ABE = \angle DAE$ , since $A$ , $C$ , $E$ lie on circle $\Gamma$ and $A$ is the point of tangency. Henceforth, $\angle DAE = \angle EDF$ .

Again, by Alternate Segment Theorem, we deduce that the circumcircle of triangle $ADE$ is tangent to line $BD$ , where $D$ is the point of tangency.

By Power of Point Theorem, $DF^2 = FE \times FA$ . On the other hand, by Power of Point Theorem, $BF^2 = FE \times FA$ . So, $DF^2 = BF^2$ , which means, $\boxed{DF = BF}$ .

Note: On the other hand, proving the parallelism of $AC$ and $BD$ might also be useful in this proof.

From above result, one can show that the length of $FB$ is half of $BD$ . But, we can see that $AD = BD$ since triangles $DAM$ and $DBM$ are congruent by RHS Property. Finally, $FB = \frac{120}{2} = \boxed{60}$ .

Let the circumcircle of $AED$ be $\Gamma_1$ . We will show that $\Gamma_1$ is tangential to $BD$ . This is true because $\angle ADB = 180^\circ - 2 \times \angle ABD = 180^\circ - 2 \times \angle BCA = \angle ABC = \angle AEC = 180^\circ - \angle DEA$ .

By power of a point, $DF^2 = FE \times FA = FB^2$ , hence $DF = FB = \frac {BD}{2} = \frac {BA}{2} = 60$ .

The intersection could be inside or outside the circle. AB and CD are chord which intersect at O (inside or outside the circle). Then OA OB=OC OD , because AB=BC then FB must be (1/2)*AD=60

Let us consider a circle x^2 + y^2 = 120 Taking A (120,0) B(0,120) and C(-120, 0), D(120,120) solving equation of CD and circle get E and solving AE and BE get coordinates of F as (60,120) Therefore distance FB = 60

Introduce an auxiliary point, $G=AB\cap CD$ .

Use the power of a point theorem of $D$ w.r.t. $\Gamma$ and similar triangles to show that $\frac{GE}{ED}=\frac{AG}{AB}$ .

Now apply Menelaus' Theorem to the hexagon $AEDFBG$ to obtain:

$\frac{AB}{AG}\cdot\frac{GE}{ED}\cdot\frac{DF}{BF}=1\implies DF=BF$

Thus, $F$ is the midpoint of $BD$ so $BF=60$ .