如图所示，两个圆交于 $D$ 和 $F$。 $ABC$ 是大圆在 $B$ 处的切线，而 $BFH$ 是小圆在 $F$ 处的切线。 $ADF$、$BDE$ 和 $GFE$ 是直线。 已知角 $BFG = x^\circ$。

1. Find the angle vertically opposite to $\angle BFG$.

Since $BFH$ and $GFE$ are straight lines intersecting at $F$,
$\angle HFE = \angle BFG = x^\circ$ (vertically opposite angles).

2. Apply the alternate segment theorem to the smaller circle.

For the smaller circle, $FH$ is a tangent at $F$ and $FE$ is a chord.
$\angle FDE = \angle HFE = x^\circ$ (alternate segment theorem).

3. Find the adjacent angle on the straight line $BDE$.

Since $BDE$ is a straight line,
$\angle FDB = 180^\circ - \angle FDE = 180^\circ - x^\circ$ (angles on a straight line).

4. Use the cyclic quadrilateral in the larger circle to find $\angle BGF$ and conclude.

Points $B, G, F, D$ lie on the larger circle, so $BGFD$ is a cyclic quadrilateral.
$\angle BGF + \angle FDB = 180^\circ$ (angles in opposite segments)
$\angle BGF + (180^\circ - x^\circ) = 180^\circ$
$\angle BGF = x^\circ$

In $\triangle BGF$, $\angle BFG = x^\circ$ (given) and $\angle BGF = x^\circ$.
Since $\angle BFG = \angle BGF$, $\triangle BGF$ is an isosceles triangle.

1. Identify angle GBF using the alternate segment theorem

Since $ABC$ is a tangent to the larger circle at $B$, by the alternate segment theorem, $\angle CBG = \angle BFG$.
Given $\angle BFG = x^\circ$, we have $\angle CBG = x^\circ$.

2. Use alternate angles to prove parallel lines

We are given that $GFE$ is a straight line, so $\angle BFG$ and $\angle CBG$ are alternate angles between lines $CBA$ and $GFE$ intersected by transversal $BG$.
Since $\angle CBG = \angle BFG = x^\circ$, the alternate angles are equal.
Therefore, $CBA$ is parallel to $GFE$.

1. First pair of equal angles using the parallel lines from (b)

From part (b), $CBA \parallel GFE$. The straight line $BDE$ is a transversal cutting these two parallel lines. Hence, by alternate angles,
$\angle ABD = \angle ABE = \angle BEG.$

2. Second pair of equal angles using vertically opposite angles, the alternate segment theorem and part (a)

Since $ADF$ and $BDE$ are straight lines crossing at $D$,
$\angle ADB = \angle EDF \quad (\text{vertically opposite angles}).$
In the smaller circle, $BFH$ is the tangent at $F$, so by the tangent–chord (alternate segment) theorem,
$\angle EDF = \angle EFH.$
Since $BFH$ and $GFE$ are straight lines crossing at $F$,
$\angle EFH = \angle BFG = x^\circ \quad (\text{vertically opposite angles}).$
From part (a), triangle $BGF$ is isosceles, so $\angle BGF = \angle BFG = x^\circ$. As $G$, $F$, $E$ are collinear, $\angle BGE = \angle BGF$. Therefore
$\angle ADB = \angle BGE.$

3. Conclude similarity

Triangles $ABD$ and $BEG$ have two pairs of equal corresponding angles:
$\angle ABD = \angle BEG, \qquad \angle ADB = \angle BGE.$
Hence triangle $ABD$ is similar to triangle $BEG$ (AA).