14. In triangle $ABC$, the difference between the internal angles $\hat{A}-\hat{B}=\varphi$ is given. It is known that the height dropped from vertex $C$ to side $AB$ is equal to the difference $|BC|-|AC|$. Find the angles of triangle $ABC$.
14. If the height $|C D|=h$, then $|B C|=\frac{h}{\sin \hat{B}},|A C|=\frac{h}{\sin \hat{A}}$. By the condition $\frac{h}{\sin \hat{B}}-\frac{h}{\sin \hat{A}}=h$, hence $\sin \hat{A}-\sin \hat{B}=\sin \hat{A} \sin \hat{B}$, $2 \sin \frac{\hat{A}-\hat{B}}{2} \cos \frac{\hat{A}+\hat{B}}{2}=\frac{1}{2}[\cos (\hat{A}-\hat{B})-\cos (\hat{A}+\hat{B})], \quad 2 \sin \frac{\varphi}{2} \times$ $\times \sin \frac{\hat{C}}{2}=\frac{1}{2}(\cos \varphi+\cos \hat{C}), 2 \sin ^{2} \frac{\hat{C}}{2}+4 \sin \frac{\varphi}{2} \sin \frac{\hat{C}}{2}-(1+\cos \varphi)=$ $=0$. This is a quadratic equation in terms of $\sin \frac{\hat{C}}{2}$ :

$$
\sin ^{2} \frac{\hat{C}}{2}+2 \sin \frac{\varphi}{2} \sin \frac{\hat{C}}{2}-\cos ^{2} \frac{\varphi}{2}=0
$$

from which

$$
\sin \frac{\hat{C}}{2}=1-\sin \frac{\varphi}{2}
$$