Junior Balkan MO 2002 - Problem 1
Dodao/la:
arhiva27. listopada 2023. 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 $|PA| + |PB| = 2 \cdot |PD|$.
Izvor: Juniorska balkanska matematička olimpijada 2002.