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 $\measuredangle EAD = \frac{53^\circ}{2}$


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
$$TC^2=16(1+\frac{4}{5}-\frac{4}{5})$$
$$TC^2=16$$
$$\boxed{TC=4}$$
