Let angle CAP=x so angle BAP=(60x) etc.
Applying the law of sines on triangles ABP and ACP gives
sin(60)/AB = sin(60+x)/AP = sin(60x)/BP and
sin(60)/AC = sin(x)/PC = sin(120x)/AP
so
BP=APsin(60x)/sin(60+x)
PC=APsin(x)/sin(120x)
these denominators are equal so
BP+PC = AP[sin(60x)+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 20130907 13:30:53 