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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

PQ is diameter . Show that QC & AP bisect angle C & angle A

*Last edited by shivusuja (2013-02-26 03:27:51)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

Hi

Need more information about this. Diagram?

Bob

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

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Yes There Is A Diagram As Follows

Abcd Is A Cyclic Quadrilateral.where Pq Is The Diameter. Show That Qc And Ap Bisect Angle C & And Angle A.

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

ABCD IS A CYCLIC QUADRILATERAL.WHERE PQ IS THE DIAMETER. SHOW THAT QC AND AP BISECT ANGLE C & AND ANGLE A.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

Hi

Also need order of all 6 points around circle

So I can place a b c d relative to p q

Did my other post answer seem ok?

Bob

*Last edited by bob bundy (2013-02-27 05:31:34)*

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

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Q Is In Between Arc AD & P Is In Between Arc BC. And PQ Is The Diameter.

This Is The Figure.

About The Previous Sum ....

Since Yesterday Iam Strugling With That. Does The Point O Comes Out Of The Circles. How Is ACD= AQB.

Iam Find ACD+AQB=180. Please Can You Give More Clarity

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

hi shivusuja

Thanks for the information.

I've made a diagram that I hope is correct, but I am unable to put it in this post. ??? I'll put it in the next instead.

No I cannot post it at all. ??? I'll work on it.

Got it. I think the image was too big as a GIF so I've re-saved it as a JPG.

But angle A is not bisected by PA (my software letters the points in alphabetical order so it has labelled point P with letter E. I put the P and Q using a text overlay.

So I'm confused about what to prove here.

Bob

ps. I'll add some more to the other post for you.

*Last edited by bob bundy (2013-02-28 07:42:03)*

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

diagram

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Hi

The Figure Is The Same Which You Drew.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

Good. But the lines do not bisect the angles. Am I mis-interpreting the question?

Maybe you can say what is to be proved in the form

angle XYZ = angle???

Bob

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

Angle BCQ= Ange QCD

&

Angle BAP=angle PAD

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,915

hi shivusuja

That's what I thought the question meant. Problem is: those angles are not equal.

I use geometry software called Sketchpad. It's a vector geometry program and can give measurements of lengths and angles, accurate to many decimal places. I've got it set to 2dp. When I select points to make measurements, the software automatically assigns letters in order (A, B C ...) so the point P has label E, the centre of the circle has label F, and the point Q has label G.

So the angles you want are shown as BCG, GCD, BAE and EAD. The screenshot below shows the Sketchpad values.

I don't know what to suggest next.

Where did the question come from? Have you worded it exactly as set?

Bob

*Last edited by bob bundy (2013-03-01 21:42:44)*

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**shivusuja****Member**- Registered: 2006-09-14
- Posts: 56

First Of All Sorry For The Late Reply. It Was From "cbse All In One Exam Practice Sums". Many Teachers From The School Also Tried The Sum But Could Not Reach The Answer. So Whenever I Stuck With Any Sum I Post It In This Site . And Definetly I Got The Solution. May Be The Sum Is Wrong.

Thanks For All.

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