# the triangle ABC has CA=CB.?

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 = 2PD.

Relevance

Here is a possible configuration:

http://farm4.static.flickr.com/3582/3827160894_7c9...

Let E is the foot of the perpendicular from C to PA. Since

arc(CA) = arc(CB), angle(CPA) = angle(CPB), the latter implies

ΔCPE ≡ ΔCPD (both right with common hypotenuse and a pair of equal acute angles).

The congruence implies |PD| = |PE|, so we'll have it proven if we prove

|PA| + |PB| = |PD| + |PE| = 2|PD|, or |PA| - |PE| = |PD| - |PB|, but

|PA| - |PE| = |AE|, |PD| - |PB| = |BD|, so it remains to prove |AE| = |BD|.

The latter segments are equal indeed, since they are legs in

ΔCAE ≡ ΔCBD (both right with equal hypotenuses - ΔABC isosceles - and angles CAP and CBP complement to 180°, angles CBP and CBD also complement to 180°, hence

angle(CAE) = angle(CBD) /white arcs on green on the picture/.

Another possible positions of P, D and E are treated similarly, what proves the required statement.