In the diagram, ABC is an equilateral triangle, ACDE is a square and angle FAC = 15°.

1. Establish one pair of equal corresponding sides

Since $ACDE$ is a square, $AE=AC$.

Since $ABC$ is an equilateral triangle, $AB=AC$.

Hence, $AE=AB$.

2. Show the included angles are equal

Since $ACDE$ is a square, $\angle EAC=90^\circ$.

Given $\angle FAC=15^\circ$,
$\angle EAF=90^\circ-15^\circ=75^\circ.$

Since $ABC$ is an equilateral triangle, $\angle BAC=60^\circ$.

Therefore
$\angle BAF=60^\circ+15^\circ=75^\circ.$

Hence, $\angle EAF=\angle BAF$.

3. Apply the congruence test

Also, $AF$ is a common side.

Therefore, in $\triangle EAF$ and $\triangle BAF$:

$AE=AB$, $AF=AF$, and $\angle EAF=\angle BAF$.

Hence, $\triangle EAF \cong \triangle BAF$ by SAS.

1. Use the angle bisector and congruence result to find an angle in triangle FAE

Since $ABC$ is equilateral, $\angle ABC=60^\circ$.

Since $BF$ bisects $\angle ABC$,
$\angle ABF=30^\circ.$

From part (a), $\triangle EAF \cong \triangle BAF$, so corresponding angles are equal:
$\angle AEF=\angle ABF=30^\circ.$

2. Show two angles in triangle FAE are equal, then conclude it is isosceles

From the square and $\angle FAC=15^\circ$,
$\angle EAF=90^\circ-15^\circ=75^\circ.$

In $\triangle FAE$,
$\angle AFE=180^\circ-75^\circ-30^\circ=75^\circ.$

Thus
$\angle AFE=\angle EAF.$

Therefore the opposite sides are equal, so $AE=EF$. Hence $\triangle FAE$ is isosceles.