In the given figure, AE = AF = AD = DE = r and arc FE = 38 arc DE. What is the value of ∠BAC?

In the given construction, the length AE = AF = AD = DE = r implies that it is a circular arc centered at A with radius r and △ADE is an equilateral triangle.
Thus, each angle of the △ADE is 60°.
∴ ∠DAB = 60°



Since arc FE = $\frac{3}{8}$ arc DE,
⇒∠BAC = $\frac{3}{8}$∠DAB
⇒∠BAC = $\frac{3}{8}$× 60°
⇒∠BAC = 22.5°
Thus, the value of ∠BAC is 22.5°.