$17 \cdot 39 P$ is a point inside an equilateral $\triangle ABC$, and the sizes of $\angle APB, \angle BPC, \angle CPA$ are in the ratio 5:6:7. Then, the size ratio of the three interior angles of the triangle formed by $PA, PB, PC$ (from smallest to largest) is 
(A) $2: 3: 4$.
(B) $3: 4: 5$.
(C) $4: 5: 6$.
(D) cannot be determined.
(China Junior High School Mathematics Correspondence Competition, 1988)
[Solution 1] As shown in Figure (1), draw $PD \parallel AB$ intersecting $BC$ at $D$, draw $PE \parallel BC$ intersecting $AC$ at $E$, and draw $PF \parallel AC$ intersecting $AB$ at $F$.
$$
\begin{array}{ll}
\text { Then } & \angle PDB = \angle PEC = \angle PFA = 120^{\circ} \\
\because & \angle APB : \angle BPC : \angle CPA = 5 : 6 : 7,
\end{array}
$$
$$
\therefore \angle APB = 100^{\circ}, \angle BPC = 120^{\circ}, \angle CPA = 140^{\circ}.
$$

And
$$
\begin{array}{l}
\angle APF + \angle PBD = \angle APB - 60^{\circ} = 40^{\circ}, \\
\angle BPD + \angle PCE = \angle BPC - 60^{\circ} = 60^{\circ}, \\
\angle CPE + \angle PAF = \angle CPA - 60^{\circ} = 80^{\circ}.
\end{array}
$$

By arranging $\triangle PFA, \triangle PDB, \triangle PCE$ as shown in Figure (2), making their $120^{\circ}$ angles coincide, we have $GH = PA, HI = PB, IG = PC$.
Thus, $\triangle GHI$ is the triangle with sides of lengths $PA, PB, PC$, where
$$
\begin{array}{l}
\angle G = \angle CPE + \angle PAF = 80^{\circ}, \\
\angle H = \angle APF + \angle PBD = 40^{\circ}, \\
\angle I = \angle BPD + \angle PCE = 60^{\circ}.
\end{array}
$$

Therefore, the ratio of the three interior angles of $\triangle GHI$ (from smallest to largest) is 2:3:4.
Hence, the answer is $(A)$.
[Solution 2] Construct an equilateral $\triangle BPD$ outside $\triangle ABC$ with $BP$ as a side, and connect $CD$ (as shown in Figure (3)).
It is easy to prove that $\triangle ABP \cong \triangle CBD$,
$$
\therefore PA = DC, \text{ and } \angle APB = \angle CDB.
$$

Since $\triangle BPD$ is an equilateral triangle, then
$$
PB = PD.
$$
$\therefore \triangle PCD$ is the triangle with sides of lengths $PA, PB, PC$.
$$
\begin{array}{l}
\because \quad \angle APB : \angle BPC : \angle CPA = 5 : 6 : 7, \text{ then } \\
\angle APB = 100^{\circ}, \quad \angle BPC = 120^{\circ}, \\
\angle CPA = 140^{\circ}. \\
\therefore \quad \angle PDC = \angle BDC - \angle BDP = \angle BPA - 60^{\circ} = 40^{\circ},
\end{array}
$$

And $\angle CPD = \angle BPC - \angle BPD = 120^{\circ} - 60^{\circ} = 60^{\circ}$,
Also, $\angle PCD = 180^{\circ} - (\angle PDC + \angle CPD) = 80^{\circ}$.
Thus, the ratio of the three interior angles of the triangle is $2:3:4$.
Hence, the answer is $(A)$.