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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,370

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,588

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)*

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,374

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

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,374

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

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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)*

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**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

@bob bundy, I was wondering how you got angle DCB = 180 - (x + y) => BCQ = x + y.

Girls can be just as good as boys at math. We just need to get the same encouragement.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,588

hi championmathgirl

Welcome to the forum.

There is always one circle that goes through the three points of any triangle. A 4th point may or may not be on this circle too. If it is the quadrilateral is called cyclic. ABCD is such a quadrilateral. All cyclic quadrilaterals have the property that opposite angles add up to 180 degrees. I proved this here:

http://www.mathisfunforum.com/viewtopic.php?id=17799 starting at post 4.

Bob

Children are not defined by school ...........The Fonz

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

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**Wangotango****Member**- Registered: Yesterday
- Posts: 4

Can someone help me find RQB, RSB. And also can someone tell me how that relates to PQR?

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 594

*Last edited by thickhead (Yesterday 19:31:34)*

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**Wangotango****Member**- Registered: Yesterday
- Posts: 4

Can you prove that the exterior angle and remote interior angle of a cylic quadrilateral is equal?

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 594

It depends on what is the starting point. If you start with the sum of opposite angles is

it is one step. Else you have to start with the angle inscribed in an arc is half the measure of arc and show that the sum of opposite angles is first.God helps those who help themselves.

There is nothing impossible on earth.

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**Wangotango****Member**- Registered: Yesterday
- Posts: 4

Ok thx I proved it, but how does RSB=RQB+D help? I'm still not sure how to finish this proof...

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 594

What exactly you need to prove? I thought you meant angle PDQ.

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**Wangotango****Member**- Registered: Yesterday
- Posts: 4

We're trying to prove that where the angle bisectors intersect is perpendicular

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 594

I thought it was proved by phrontister. Any way I have have given the proof in some recent thread. My current response was just for #10.

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 594

Hi Wangotango,

Please see the following thread

http://www.mathisfunforum.com/viewtopic.php?id=23185

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There is nothing impossible on earth.

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