Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 62

Let ABCD be a cyclic quadrilateral. Let P be the intersection of \overline{AD} and \overline{BC}, and let Q be the intersection of \overline{AB} and \overline{CD}. Prove that the angle bisectors of \angle DPC and \angle AQD are perpendicular.

I just don't know how to get this one. This problem in one word=aggggggghhhhhh.

would power of a point help?

thanks!

Batman shows Superman his new phone. Batman-"Alfred, whats the temperature in London?"

Alfred-"Just a second, sir." Alfred turns on his iPhone."Siri, whats the temperature in London?"

Siri - "Six centigrade." Alfred-"Six centigrade sir." Batman-"See?"

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,652

Hi;

Hi; I put the image in for you. Also you need to use the math tags for your latex.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,119

hi thedarktiger,

This time I'm more confident about what to do. (But I haven't fully tried it yet. )

Call angle APB = x and ABP = y

So DAB = x + y => DCB = 180 - (x + y) => BCQ = x + y

So get an expression in terms of x and y for BQC.

Let the bisectors cross at R, and let RQ cross BC at S

Get an expression in terms of x and y for RPB and RQB, then RSB.

Hence for PRQ.

EDIT: Thought I'd better check it ..... it does work.

Bob

*Last edited by bob bundy (2014-01-28 00:02:42)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Hi,

Drew it up in Geogebra, which gave the two bisectors as being perpendicular to each other.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 62

wow! thanks! @bob bundy, you are really good at geometry. phrostister and bobym, thanks for the pics!

Batman shows Superman his new phone. Batman-"Alfred, whats the temperature in London?"

Alfred-"Just a second, sir." Alfred turns on his iPhone."Siri, whats the temperature in London?"

Siri - "Six centigrade." Alfred-"Six centigrade sir." Batman-"See?"

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,811

Here is a little video of post #4, demonstrating the perpendicular property of the two bisectors in various cyclic quadrilateral shapes. Done in Geogebra by randomly dragging the points of the quadrilateral along the curve of the circle.

*Last edited by phrontister (2014-01-29 23:44:45)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 62

thanks a lot for the video! So I guess they are always perpendicular.

I think there should be some kind of theorem ... maybe the bob-tister-bob theorem of secants.

Maybe I will make a wikipedia page.

*Last edited by thedarktiger (2014-02-02 21:56:00)*

Batman shows Superman his new phone. Batman-"Alfred, whats the temperature in London?"

Alfred-"Just a second, sir." Alfred turns on his iPhone."Siri, whats the temperature in London?"

Siri - "Six centigrade." Alfred-"Six centigrade sir." Batman-"See?"

Offline

Pages: **1**