What is the measurement of the $CT$ segment in the square below? For reference The side of square $ABCD$ measures $4$. Calculate TC. (answer: 4)

My progress:
Por propriedade BTE é $90 ^ \circ$
G is the barycenter of the $\triangle ADC$
We have a 2:4 ratio in the $\triangle DEA$ so $\measuredangle EAD = \frac{53^\circ}{2}$

 A: Without (many) words, $TC$ is the median to the hypotenuse in right triangle $TBB'\,$.

A: Without much thoughts:
Let $A = (0,0)$, line $BF: y = -\frac12 x + 2$, line $AE: y = 2x$, then $T = (0.8, 1.6)$.
Length $CT = \sqrt{(4-0.8)^2+(4-1.6)^2} = 4$
A: Let's say $\angle ABF=\theta$. It's not hard to see that
$$BT=4\cos\theta$$
$$\angle TBC=90-\theta$$
By Law of Cosines on $\angle TBC$ in $\Delta TBC$, we have
$$TC^2=BT^2+BC^2-2\cdot BT\cdot BC\cos\angle TBC$$
$$TC^2=16\cos^2\theta+16-32\cos\theta\sin\theta$$
$$TC^2=16(1+\cos^2\theta-2\cos\theta\sin\theta)$$
Since $\sin\theta=\frac{1}{\sqrt{5}}$ and $\cos\theta=\frac{2}{\sqrt{5}}$, we have
$$TC^2=16(1+\frac{4}{5}-\frac{4}{5})$$
$$TC^2=16$$
$$\boxed{TC=4}$$
A: Construct the line through $C$ parallel to $AE$, intersecting $AB$ at its midpoint and $BT$ at $M$, which is then the midpoint of $BT$.
Hence $CM$ is both median and altitude of triangle $CBT$, which is then isosceles: the result follows.
A: I would use an algebraic method.  Set up a coordinate system so that A is at (0, 0), B is at (1, 0), C is at (1, 1) and D is at (0, 1).   Then E is at (1/2, 1) and the line AE is y= 2x.  F is at (0, 1/2) and the line BF is y= (-1/2)x+ 1/2
The two lines intersect where y= 2x= (-1/2)x+ 1/2 so x= 1/5, y= 2/5. T is (1/5, 2/5) and the distance from T to C is $\sqrt{(1- 1/5)^2+ (1- 2/5)^2}= \sqrt{(4/5)^2+ (3/5)^2}= \sqrt{16/25+ 9/25}= \sqrt{25/25}= 1$.
Since I belatedly see that we were actually given that the square has side length 4, the answer is actually 4 times that, 4(1)= 4.
