Construct a regular pentagon

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I want to construct a regular pentagon with side length $4$.

I have done the following:

We are given the segment $|AT|=4$ and we will create a point $B$ outside of $|AT|$ such that $|AB|$ is divided by $T$ in golden ratio.

We create the perpendicular line segment to $|AT|$ from $T$ with length $\overline{AT}$ and so we get the point $C$.

Then we create the midpoint $M$ of $|AT|$.

The circle around $M$ with radius $\overline{MC}$ intersects the extension of $|AT|$ from $T$ at the point $B$.

So we have that $\frac{\overline{AB}}{\overline{AT}}=\Phi$.

From that equation we get that $\overline{AB}=\overline{AT}\cdot \Phi=4\Phi$.

Then we create an isosceles triangle on the segment $|AT|$ with the two equal sides of length $\overline{AB}$ as follows:

We create a circle with center $A$ and radius $\overline{AB}$ and a circle with center $T$ and radius $\overline{AB}$. Let $P$ be the intersection point of the two circles, above of $|AB|$. This is a triangle of type $1$.

On the sides $|AP|$ and $|TP|$ we create an isosceles triangle with two equal sides of length $4$.

Let's consider first the side $|AP|$. We create a circle with center $A$ and radius $4$ and a circle with center $P$ and radius $4$. Let $K_1$ be the intersection point of the two circles, above of $|AP|$. This is a triangle of type $2$.

Now we consider the side $|TP|$. We create a circle with center $T$ and radius $4$ and a circle with center $P$ and radius $4$. Let $K_2$ be the intersection point of the two circles, above of $|TP|$. This is a triangle of type $2$.

Now we have a regular pentagon.

Is everything correct?

To justify that these are indeed triangles of type $1$ and $2$ respectively do we calculate the angles using the cosine rule?