This exercise is adapted from a problem proposed by Professor Norman Schaumberger in the May 1990 issue of The College Mathematics Journal. In the accompanying figure, radius $O A=1$ and $\angle A F C=60^{\circ} .$ Follow steps (a) through (f) to prove that $$A D \cdot A B-A E \cdot A C=B C$$
(a) Let $\angle B A F=\alpha ;$ show that $A D=2 \cos \alpha$
Hint: In isosceles triangle $A O D,$ drop a perpendicular from $O$ to side $\overline{A D}$.
(b) Let $\angle C A F=\beta ;$ show that $A E=2 \cos \beta$
(c) Explain why $\angle B=60^{\circ}-\alpha$ and $\angle C=120^{\circ}-\beta$
(d) Using the law of sines, show that
$$ A B=\frac{B C \sin \left(120^{\circ}-\beta\right)}{\sin (\alpha+\beta)} $$
and
$$ A C=\frac{B C \sin \left(60^{\circ}-\alpha\right)}{\sin (\alpha+\beta)} $$
(e) Using the results in parts (a), (b), and (d), verify that $A D \cdot A B-A E \cdot A C$
$$=B C \cdot\left[\frac{2 \cos \alpha \sin \left(120^{\circ}-\beta\right)-2 \cos \beta \sin \left(60^{\circ}-\alpha\right)}{\sin (\alpha+\beta)}\right]$$
(f) To complete the proof, you need to show that the quantity in brackets in part (e) is equal to $1 .$ In other words, you want to show that
$$\begin{aligned} 2 \cos \alpha \sin \left(120^{\circ}-\beta\right)-2 \cos \beta \sin \left(60^{\circ}\right.&-\alpha) \\ &=\sin (\alpha+\beta) \end{aligned} $$
Use the addition formulas for sine to prove that this last equation is indeed an identity.