1. In the tetrahedron $ABCD$, the height from vertex $D$ falls inside $\triangle ABC$, and the faces $\triangle DAB$, $\triangle DAC$, and $\triangle DBC$ form equal dihedral angles with the face $\triangle ABC$. Given that $P(\triangle ABC)=21$, $P(\triangle DAB)=15$, $P(\triangle DAC)=13$, and $P(\triangle DBC)=14$, calculate the volume of the tetrahedron $ABCD$.
1. Let $S$ be the foot of the altitude from $D$. Let $A_{1}, B_{1}$, and $C_{1}$ be the feet of the perpendiculars from $S$ to $B C, A C$, and $A B$, respectively. According to the theorem of the three perpendiculars, $D A_{1}, D B_{1}$, and $D C_{1}$ are perpendicular to $B C, A C$, and $A B$, respectively. Therefore, $\measuredangle D A_{1} S, \measuredangle D B_{1} S$, and $\measuredangle D C_{1} S$ are dihedral angles with edges $B C, A C$, and $A B$, and they are equal to each other by the problem's condition; let's denote them by $\theta$. Now, by the ASA (Angle-Side-Angle) theorem, we have $\triangle D S A_{1} \cong \triangle D S B_{1} \cong \triangle D S C_{1}$ (all have the common side $D S$, one right angle, and one angle $\theta$). From this, it follows that $S A_{1}=S B_{1}=$

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Op 2019 4A 1 $S C_{1}=r$ (i.e., $S$ is the center of the inscribed circle of $\triangle A B C$, and $S$ is inside $\triangle A B C$), and $D A_{1}=D B_{1}=D C_{1}=h$, and also $\frac{r}{h}=\cos \theta$. Now we have $\frac{P(\triangle S A B)}{P(\triangle D A B)}=\frac{P(\triangle S A C)}{P(\triangle D A C)}=\frac{P(\triangle S B C)}{P(\triangle D B C)}=\frac{r}{h}=\cos \theta$ (since $r$ and $h$ are, respectively, the heights of the corresponding triangles to $A B, A C$, and $B C$), so it follows that $P(\triangle S A B)=15 \cos \theta, P(\triangle S A C)=13 \cos \theta$, and $P(\triangle S B C)=14 \cos \theta$, and from there $21=P(\triangle A B C)=P(\triangle S A B)+P(\triangle S A C)+P(\triangle S B C)=15 \cos \theta+13 \cos \theta+14 \cos \theta=$ $42 \cos \theta$. We get $\cos \theta=\frac{1}{2}$, i.e., $\theta=60^{\circ}$, and $P(\triangle S A B)=\frac{15}{2}, P(\triangle S A C)=\frac{13}{2}$, and $P(\triangle S B C)=7$.

From the above relations, it follows that $A B: A C: B C=15: 13: 14$ (since $\triangle S A B$, $\triangle S A C$, and $\triangle S B C$ all have the height $r$, and their areas are in the given proportion). Let $A B=15 x, A C=13 x$, and $B C=14 x$. From Heron's formula, we have $21=P(\triangle A B C)=\sqrt{21 x \cdot 7 x \cdot 8 x \cdot 6 x}=84 x^{2}$, from which it follows that $x=\frac{1}{2}$, i.e., for example, $A B=\frac{15}{2}$. Now we have $h=\frac{2 P(\triangle D A B)}{A B}=\frac{30}{\frac{15}{2}}=4$, and then from $\triangle D S C_{1}$ we get $D S=h \sin \theta=4 \sin 60^{\circ}=2 \sqrt{3}$. Finally, the volume of the tetrahedron is: $V=\frac{D S \cdot P(\triangle A B C)}{3}=\frac{2 \sqrt{3} \cdot 21}{3}=14 \sqrt{3}$.