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Topic: Without using the cosine rule or steps from its derivation...
Replies: 15   Last Post: May 4, 2014 10:41 AM

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 quasi Posts: 12,067 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

Date Subject Author
4/29/14 Port563
4/29/14 Brian Q. Hutchings
4/29/14 quasi
4/29/14 Port563
4/29/14 quasi
4/29/14 Ken.Pledger@vuw.ac.nz
4/30/14 quasi
4/30/14 quasi
4/30/14 Leon Aigret
4/30/14 Port563
4/30/14 Ken.Pledger@vuw.ac.nz
5/1/14 Port563
5/2/14 sharma.kunapalli@gmail.com
5/2/14 quasi
5/4/14 Port563
5/4/14 Leon Aigret