1998
Problem - 3120
In $\triangle{ABC}$, $\angle{BAC} = 40^\circ$ and $\angle{ABC} = 60^\circ$. Points $D$ and $E$ are on sides $AC$ and $AB$, respectively, such that $\angle{DBC}=40^\circ$ and $\angle{ECB}=70^\circ$. Let $F$ be the intersection point of $BD$ and $CE$. Show that $AF\perp BC$.
It is easy to show that (see %%HREF%%3308%%)
\begin{equation}
AF\perp BC\Leftrightarrow AB^2 - AC^2 = FB^2 - FC^2
\end{equation}
Therefore all need to do is to verify the above relation by transforming all these line segments using trigonometric expressions and then evaluating.
Let $BC=1$, then by law of sines:
$$AB = \sin\angle{ACB}\cdot\frac{BC}{\sin\angle{BAC}} = \frac{\sin{80^\circ}}{\sin{40^\circ}}$$
$$AC = \sin\angle{ABC}\cdot\frac{BC}{\sin\angle{BAC}} = \frac{\sin{60^\circ}}{\sin{40^\circ}}$$
$$FB = \sin\angle{FCB}\cdot\frac{BC}{\sin\angle{BFC}} = \frac{\sin{70^\circ}}{\sin{70^\circ}}$$
$$FC = \sin\angle{FBC}\cdot\frac{BC}{\sin\angle{BFC}} = \frac{\sin{40^\circ}}{\sin{70^\circ}}$$
Then
\begin{align*}
&AB^2-AC^2\\
=&\frac{\sin^280^\circ-\sin^260^\circ}{\sin^240^\circ}\\
=&\frac{\frac{1}{2}((1-\cos 160^\circ)-(1-\cos 120^\circ))}{\sin^2 40^\circ}\\
=&\frac{\frac{1}{2}(\cos 120^\circ-\cos 160^\circ)}{\sin^2 40^\circ}\\
=&\frac{\sin 140^\circ \sin 20^\circ}{\sin^2 40^\circ}\\
=&\frac{\sin 20^\circ}{\sin 40^\circ}\\
=&\boxed{\frac{1}{2\cos 20^\circ}}
\end{align*}
Similar, we can compute $FB^2 - FC^2$ as following:
\begin{align*}
&FB^2 - FC^2\\
=&\frac{\sin^2 70^\circ-\sin^2 40^\circ}{\sin^2 70^\circ}\\
=&\frac{\frac{1}{2}((1-\cos 140^\circ)-(1-\cos 80^\circ))}{\sin^270^\circ}\\
=&\frac{\frac{1}{2}(\cos 80^\circ-\cos 140^\circ)}{\sin^270^\circ}\\
=&\frac{\sin 110^\circ \sin 30^\circ}{\sin^2 70^\circ}\\
=&\boxed{\frac{1}{2\sin 70^\circ}}
\end{align*}
Hence $$\cos 20^\circ = \sin 70^\circ \implies AB^2-AC^2 = FB^2 - FC^2$$