19.43 If the sum of the measures of angles $A$, $B$, $C$, $D$, $E$, and $F$ in the figure on the right is $90 n^{\circ}$, then $n$ equals
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(E) 6.
[Solution 1] Let $P, Q$ be the intersections of $A D$ and $B F, C E$, respectively, then we have
$$
\begin{array}{l}
\angle F + \angle F P D + \angle E Q A + \angle E = 360^{\circ}, \\
\angle B + (180^{\circ} - \angle F P D) + \angle D = 180^{\circ}, \\
\angle A + (180^{\circ} - \angle E Q A) + \angle C = 180^{\circ}.
\end{array}
$$

Adding these equations, we get
$$
\begin{array}{l}
\angle A + \angle B + \angle C + \angle D + \angle E + \\
\angle F + 360^{\circ} = 720^{\circ}.
\end{array}
$$

The required sum is $360^{\circ} = 90 n^{\circ}$,
thus $n = 4$.
Therefore, the answer is (C).
[Solution 2] Similar to problem 19.38 [Solution 2].