The triangle $ABC$ has $|CA| = |CB|$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $\overline{AB}$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $|PA| + |PB| = 2 \cdot |PD|$.
Školjka 
has 
. 
is a point on the circumcircle between 
and 
(and on the opposite side of the line 
to 
). 
is the foot of the perpendicular from 
. Show that 
.