Circle tangent triangle perimeter

$DE = AE$ and $DF = BF$ , because tangents drawn from a single point are equal in length. Therefore, we can see that $EF = AE + BF$ .

$AOBC$ forms a square, with $O$ being the centre of the circle $\Gamma$ , so $CB = AC = 123$ .

We can see that:

$CE = 123 - AE$

$CF = 123 - BF$

Therefore the perimeter of $CEF$ is:

$CE + EF + CF = 123 - AE + 123 - BF + AE + BF = \fbox{246}$

We can think the point D is on the point A or B. So the perimeter will be just double as third side =0

well,I considered a special case here thats why i reached the answer quickly.the guy who knows its general solution is welcomed here to write it

Since $C$ is the intersection of the tangents, $AB = AC$ . Further if $O$ is the center of the circle then $AO = OB = 123$ Since $AC \perp BC$ and $AO \perp AC \Rightarrow AO \parallel BC$ . One can also show that in fact $AO = BO = AC = CB$

Next the point $E$ lying on $AC$ is the point of intersection of the tangents $AE$ and $ED$ . Thus $DE = AE$ . And similarly $DF = FB$

Next the $Perimeter(CEF) = CE + EF + CF$ , where $EF = DE +DF = AE + BF$

$\Rightarrow CE = AC - AE, CF = CB - BF \Rightarrow Perimeter(CEF) = AC + BC = 246$