It's very easy problem. Let R is radius of circle then we have that: $ RHS = 4R^{2} \sin\angle ABC \cdot \sin\angle BAC=LHS $

Firstly note that $A,D,C,E$ and $B,D,C,F$ are concyclic. Also $\angle CAE=\angle CBA, \angle CBF=\angle CAB$. Now apply sine-law in $\triangle CDE$ to get $\dfrac{CE}{CD}=\dfrac{\sin \angle CDE}{\sin \angle CED}=\dfrac{\sin \angle CAE}{\sin \angle CAD}=\dfrac{\sin \angle CBA}{\sin \angle CAB}$. Similarly we can get, by applying sine-law in $\triangle CDF$, that $\dfrac{CD}{CF}=\dfrac{\sin \angle CAB}{\sin \angle CBA}$. So $\dfrac{CE}{CD}=\dfrac{CD}{CF}$. Thus we are done.

How is it possible that such an easy problem is given at TST? O.o

I think it was a bad choice to choose this problem because everyone did it and nobody can have some upper adventage on that problem because it is too easy. Anyway I think tomorrow, quality of problems will be better and so will be harder.

An easy one: $\triangle CDB ~\triangle CEA \Longrightarrow \frac{CD}{CE} = \frac{BC}{AC}$ $\triangle CDA ~\triangle CFB \Longrightarrow \frac{CD}{CF} = \frac{AC}{BC}$ So mutiplying both these , we get $CD^2 = CE.CF$ Hence proved

there is two inscribed quadrilaterals ADCE and BDCF and two tangents AE and BF then "angle DFC" "equal" "angle DBC" "equal" "angle AEC" "equal" "angle ADC"

$CD=\sqrt{CE \cdot CF}$ is eqivalent to $CD^2=CE \cdot CF$ and because $\angle ECD=\angle FCD$ it is equivalent to $\triangle CDF \sim \triangle CED$ which is again equivalent to $\triangle ADE \sim \triangle BFD$ which is equivalent to $\frac{AD}{BF}=\frac{AE}{BD}$.Now we note that $\triangle ADC \sim \triangle BFC$ so we obtain $\frac{AD}{BF}=\frac{AC}{BC}$.Also note that $\triangle AEC \sim \triangle BDC$ so from here we have $\frac{AE}{BD}=\frac{AC}{BC}$ so the task is proven.

insta-solved without paper: Because $ \angle EAD = \angle FBD $ and $ \angle CEA =\angle CFB $, we have $ CDAE \sim DCFB \Rightarrow \frac{CD}{CF}=\frac{CE}{CD} \Rightarrow CD=\sqrt{CE\cdot CF} $

From casy's theorem we have $(\frac{CE}{CA})^2=(\frac{CF}{CB})^2=\frac{1}{2R}$ and also we have $CA=2R \sin \angle ABC$ And $CD=BC. \sin ABC$ the remaining is trivial.

Let $CE=AC \sin x $, $CD=BC\sin x $, $CF=BC\sin y $ and $\frac {AC}{\sin x} $ $=$ $\frac {BC}{\sin y} $ $\implies $ $CD^2$ $=$ $BC^2 \sin^2 x $ $=$ $AC \sin x BC \sin y $

First of all denote the angles $\angle ABC = \alpha $,and $\angle CAB = \beta$ The condition can be rewritten as $CD^2 = CE.CF \implies \frac{CD}{CE} = \frac{CF}{CD}$ So in a sense we need to prove this relation. The quads $AECD$,$CFBD$ are cyclic. A quick angle chase: $\beta = \angle ABC = \angle EAC = \angle EDC $, $\beta = \angle DBC = \angle DFC $ $\alpha = \angle CAB = \angle CBF = \angle CDF $,$\alpha = \angle CAB = \angle CAD = \angle DEC$ Thus we have that: $\triangle CDF \sim \triangle CED$ Thus we have that $\frac{CD}{CE} = \frac{CF}{CD}$,which is what we needed to prove....