Taiwan IMO Selection Camp

由於 $BC$ 是圓 $O_1$ 的直徑, $ACD$ 是 $O_1$ 的切線, 所以 $BC \perp AD$, $\angle ACB = 90^\circ$。故 $AB$ 為圓 $O_2$ 的直徑。
因為 $\angle BEC = 90^\circ$, 所以 $AB \perp CF$, 故 $\angle FAB = \angle CAB$。

連 $CG$, 知 $CG \perp BD$。故
$$\angle ADB = \angle BCG = \angle BEG = \angle AEH,$$
所以 $\triangle AHE \sim \triangle ABD$。於是有 $\frac{AH}{AE} = \frac{AB}{AD}$, 即 $AH \cdot AD = AE \cdot AB$。
由圓幂定理知 $AC^2 = AE \cdot AB$, 故 $AH \cdot AD = AE \cdot AB = AC^2 = AC \cdot AF$。所以 $\frac{AH}{AF} = \frac{AC}{AD}$, 再由合分比知 $\frac{AH}{HF} = \frac{AC}{CD}$, 證畢。

Since $BC$ is the diameter of circle $O_1$, and $ACD$ is a tangent line of $O_1$, we have $BC \perp AD$, $\angle ACB = 90^\circ$. Hence $AB$ is the diameter of circle $O_2$.
Because $\angle BEC = 90^\circ$, we have $AB \perp CF$, hence $\angle FAB = \angle CAB$.

Join $CG$, and we know $CG \perp BD$. Hence
$$\angle ADB = \angle BCG = \angle BEG = \angle AEH,$$
so $\triangle AHE \sim \triangle ABD$. Thus we have $\frac{AH}{AE} = \frac{AB}{AD}$, that is, $AH \cdot AD = AE \cdot AB$.
By the power of a point theorem, $AC^2 = AE \cdot AB$, hence $AH \cdot AD = AE \cdot AB = AC^2 = AC \cdot AF$. Therefore $\frac{AH}{AF} = \frac{AC}{AD}$, and then by the property of composition of ratios, $\frac{AH}{HF} = \frac{AC}{CD}$. This completes the proof.