Πέμπτη, 12 Οκτωβρίου 2017

2002 JBMO problem 1 (GRE)

proposed by Greece

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 ${AB }$  to ${C}$). ${D }$  is the foot of the perpendicular from ${C }$  to ${PB}$. Show that ${PA + PB = 2 \cdot PD }$.

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