geometry problem

[fiery123]

P is a point outside the circle.
A, B and P lie on one line.
C, D and P lie on a line.
A, B, C and D are cyclic as shown.
Show that PB X PA = PC X PD

[opus]

Is that all?
I hink you missed some setting, do the connecting-lines AC and BD cross at a specific angle like 90°)? ............?
And what is PB X PB? Lenth of Line PA multiplied by Length of Line PB?

[fiery123]

no there arent any extra information given. It might be that there may be a mistake in the book and missed out some information, although unlikely. I dont know what PB X PB is, but it shouldnt be PA X PB as it did not say that PB = PA.

[eranga262154]

as opus said I think you missed some data on this question. without any important data how we solved it for you. your figuer is just with few lines. so check your note agin. without any angle or lenght of a side can't say anything. there is no mean PB X PA without those data.

[triggernum5]

ABP is a Line
CDP is a Line
Therefore
BC || AD -- OAT
ADP ~ BCP -- AAA
<BP = <CP -- ZANGLE EQV + CAT
AB/sin(<AB) = DC/sin(<DC) -- SineLaw

<AB = <DC -- OAT
Therefore
BP = CP
AB = DC
ABXBP = DCXCP

[opus]

If I'm not mistaken that is only valid if the centre of the circle lies on a line that crosses the point P exactly halfway between the two other lines.

[triggernum5]

That must be the case however since 4 points on ADP lie on the circle and extend to a common point.. Could have sworn I addressed that, but I guess if I did it was poorly done.. Its been a LONG time since I've done this stuff.. Elegantly show isocoles relation and its done..

[opus]

I was thinking the same way, that was the reason for asking:do the connecting-lines AC and BD cross at a specific angle like 90°
Which would result in BC || AD.

[triggernum5]

They don't need to cross at 90deg.. Since AB is a segment of AP and DC is a segment of DP and ADBC all lie on the circle. Info on that angle isn't necessary since B and C are on AP and DP.. If I get a few more free minutes I'll try to prove it properly..
Needless to say,
AB/sin(<AB) = DC/sin(<DC)
AB/sin(<DC) = DC/sin(<AB)
sin(<AB) = sin(<DC)
Therefore
AB = DC
Triangles ABO = DCO (Unlabeled Intersection = O)

Geez, the phone rings everytime elegance pops into my head, but the OAT/CAT proves that OBC and OAD and therefore PAD are all isocoles though..

[triggernum5]

If I'm not mistaken that is only valid if the centre of the circle lies on a line that crosses the point P exactly halfway between the two other lines.

Rereading this made me notice one thing.. This problem directly demonstrates the method for bisecting an angle with a compass..:)

[fiery123]

well i suppose you are right triggernum5.
But i dont get this line - <BP = <CP -- ZANGLE EQV + CAT
What does ZANGLE EQV + CAT means? I havent came across it before.

[triggernum5]

You know the rule that says the oposite inside corners of a line crossing parallel lines in a Z looking shape, or C looking shape, and angle equality.. I call that ZANGLE EQV, and CAT is just complimentary angle theorem.. I know my logic is right here, but the proof would get part marks at best..

[fiery123]

ok i get it now.
<PAD = <PBC corr.<s BC//AD
<PAD = <PCB ext.< cyclic quad
so <PBC = <PCB which is the same as <BP = <CP
The rest i understand so thanks for the help.!