Problem 642

Solution

Since the line $t$ is tangent to the circle, we have $\angle X A E=\angle A C B$. Furthermore, since $t$ and $E F$ are parallel, we have $\angle X A E=\angle A E F$. Similarly, we have $\angle Y A F=\angle A F E=\angle A B C$. Therefore, $\triangle A F E \simeq \triangle A B C$. Thus,

$$\begin{aligned} \frac{A E}{A C} & =\frac{A F}{A B} \\ \frac{12}{10+14} & =\frac{10}{12+E B} \\ 12+E B & =20 \\ E B & =8 \end{aligned}$$

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