quasi
Posts:
11,911
Registered:
7/15/05


Re: Without using the cosine rule or steps from its derivation...
Posted:
May 2, 2014 10:39 PM


snmp wrote: > >What is the solution to this problem as stated by the OP?
I'll restate the problem, modified slightly, as follows:
Prove:
If convex cyclic pentagon PABCD is such that angles APB,BPC,CPD are equal, then (PB + PD)/(PA + PC) = PC/PB.
Proof (ignoring restrictions on allowable methods):
Inscribe pentagon PABCD in a circle.
Since angles APB,BPC,CPD are equal, chords AB,BC,CD have equal length, equal to v, say. Thus
AB = BC = CD = v
Since angles APC and BPD are equal, chords AC,BD have equal length, equal to w, say. Thus
AC = BD = w
Quadrilaterals PBCD and PABC are convex cyclic, hence Ptolemy's Theorem yields
v*PB + v*PD = w*PC
v*PA + v*PC = w*PB
Dividing the equations, we get
(PB + PD)/(PA + PC) = PC/PB
as was to be shown.
quasi

