Problem 948

Proof: $\because$ Point $V$ is on the perpendicular bisector of line segment $AB$,
$$\therefore \angle VAB = \angle VBA = \alpha \text{.}$$
$\because \angle A$ is the smallest interior angle of $\triangle ABC$, and $\alpha < \angle CAB$,
$\therefore \alpha < \angle CBA$. That is, $V$ is inside $\angle ABC$.
Similarly, $W$ is inside $\angle ACB$.
Below, we use $A, B, C$ to represent $\angle CAB, \angle ABC, \angle BCA$ respectively.
$\because W$ is on the perpendicular bisector of line segment $AC$, then
$$\begin{array}{l} \angle 1 = \angle 2 = A - \alpha, \\ \angle 3 = C - \angle 1 = C + \alpha - A \\ \therefore \angle 4 = 180^\circ - \angle 5 - \angle 3 \\ \quad = 180^\circ - (B - \alpha) - (C + \alpha - A) = 2A. \end{array}$$
In $\triangle BCT$, $\frac{BC}{\sin \angle 4} = \frac{BT}{\sin \angle 3} = \frac{CT}{\sin \angle 5}$.
$$\begin{array}{l} \because B - \alpha = 180^\circ - (C + \alpha + A), \\ \therefore TB + TC = \frac{BC}{\sin \angle 4} (\sin \angle 3 + \sin \angle 5) \\ \quad = \frac{BC}{\sin 2A} [\sin (C + \alpha - A) + \sin (B - \\ \alpha)] \\ = \frac{BC}{\sin 2A} [\sin (C + \alpha - A) + \sin (C + \alpha \\ \quad + A)] \\ = \frac{BC}{\sin 2A} \cdot 2 \sin (C + A) \cos A \\ = \frac{BC}{\sin A} \cdot \sin (C + \alpha) \\ = 2R \sin (C + \alpha). \end{array}$$

Connecting $CU, \angle BCU = \sim$.
Then $\angle \triangle CU = C + \alpha$.
In $\triangle ACJ$,
$$\begin{array}{l} AU = 2R \sin (C + \alpha). \\ \therefore AU = TB + TC. \end{array}$$