Equilateral triangle in a regular pentagon

Question

I don't understand how the line segment of $|BF|$ equals to $|BC| = |FC| = |AB| = |AE|$. How can the line segment maintains the equilateral triangle? The instructor says while drawing that it just maintains a equilateral triangle, but not how.

Answer 1

$|BC|=|AE|$ because it's a regular pentagon; then $|BC|=|FC|$ by transitivity of equality. $\angle FCB$ is $60^{\circ}$ by subtracting $48^\circ$ from $108^\circ$; now you have SAS on the triangle $\triangle BCF$, so the rest of the information is uniquely determined. Since an equilateral triangle obviously has two equal sides and a $60^{\circ}$ angle, it must be the unique triangle fitting this SAS relation. But note that $\triangle BAF$ isn't equilateral; despite how the diagram is drawn, $ABCF$ isn't an equilateral rhombus.