Point P lies on the circumcircle of equilateral triangle ABC in the
interior of ∠BAC.

Prove that |AP| = |BP| + |CP|.

Let angle CAP=x so angle BAP=(60-x) etc.
Applying the law of sines on triangles ABP and ACP gives
sin(60)/AB = sin(60+x)/AP = sin(60-x)/BP and
sin(60)/AC = sin(x)/PC = sin(120-x)/AP
so
BP=APsin(60-x)/sin(60+x)
PC=APsin(x)/sin(120-x)
these denominators are equal so
BP+PC = AP[sin(60-x)+sin(x)]/sin(60+x)
if you apply the angle sum identity the part in the brackets is shown to equal the denominator so that
BP+PC=AP

Posted by Jer on 2013-09-07 13:30:53