在圓 $CDE$ 加點 $F$。連結 $DF$ 及 $EF$。
考慮 $\Delta CDF$。
$\begin{array}{rcll}
\angle CDF & = & \angle BCQ & \text{(內錯弓形的圓周角)} \\
\angle CDF & = & 35^\circ
\end{array}$
由此，可得
$\begin{array}{rcll}
\angle ADE + \angle EDF + \angle CDF & = & 180^\circ & \text{(直線上的鄰角)} \\
100^\circ + \angle EDF + 35^\circ & = & 180^\circ \\
\angle EDF & = & 45^\circ
\end{array}$
考慮 $\Delta ABC$。
$\begin{array}{rcll}
\angle BAC & = & \angle BCQ & \text{(內錯弓形的圓周角)} \\
\angle BAC & = & 35^\circ
\end{array}$
考慮 $\Delta ADE$。
$\begin{array}{rcll}
\angle ADE + \angle AED + \angle DAE & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
100^\circ + \angle AED + 35^\circ & = & 180^\circ \\
\angle AED & = & 45^\circ
\end{array}$
考慮 $\Delta DEF$。
$\begin{array}{rcll}
\angle DFE & = & \angle AED & \text{(內錯弓形的圓周角)} \\
\angle DFE & = & 45^\circ
\end{array}$
由此，可得
$\begin{array}{rcll}
\angle DEF + \angle EDF + \angle EFD & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
\angle DEF + 45^\circ + 45^\circ & = & 180^\circ \\
\angle DEF & = & 90^\circ
\end{array}$
考慮四邊形 $CDEF$。
$\begin{array}{rcll}
\angle DCF + \angle DEF & = & 180^\circ & \text{(圓內接四邊形的對角)} \\
\angle DCF + 90^\circ & = & 180^\circ \\
\angle DCF & = & 90^\circ
\end{array}$
由此，考慮 $\Delta ABC$，可得
$\begin{array}{rcll}
\angle ABC + \angle ACB + \angle BAC & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
\angle ABC + 90^\circ + 35^\circ & = & 180^\circ \\
\angle ABC & = & 55^\circ
\end{array}$
