6. $P$ is a point inside Rt $\triangle A B C$, $\angle B=90^{\circ}, P A=10, P B=6, \angle A P B=$ $\angle B P C=\angle C P A, P C=$ $\qquad$
6. 33 .
$\angle A P B=\angle B P C=\angle C P A=\frac{360^{\circ}}{3}=120^{\circ}$, in $\triangle A B P, \triangle B C P, \triangle A C P$ using the cosine rule we have $A B^{2}=A P^{2}+B P^{2}-2 A P \cdot B P \cdot \cos 120^{\circ}=196, B C^{2}=36+P C^{2}+6 P C, C A^{2}=$ $P C^{2}+100+10 P C$. Substituting the above three equations into $A B^{2}+B C^{2}=A C^{2}$, we get $196+36+P C^{2}+6 P C=P C^{2}+$ $100+10 P C, P C=33$.