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**Prove this**

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ABCD is a square

PQ = QC = DR

Prove that: angle QPR = 45 degrees

Caution: I am not sure if I have drawn the following diagram perfectly or correctly

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,897

Hi Agnishom;

Are you sure it is DR?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,009

hi Agnishom,

Hhhmm. as given, that diagram is impossible.

DR > DC > QC, so DR = QC is impossible.

Bob

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

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How about it being BR?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,009

I think that might be possible. I'll get to work on it.

LATER I've used Sketchpad to make an accurate construction and QPR = 46.13

So that's not it I'm afraid.

EVEN LATER:

I've made a construction for the square, the property PQ = QC and QPR = 45.

When I use Sketchpad to measure angles and distances the software puts in letter labels automatically, so I've ended up with F,G and H rather than P, Q and R. But the diagram has the right features apart from that. I cannot find a third length equal to PQ and QC.

Bob

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,897

Hi;

But if there are lots of points that have the condition

QC = PQ = BR each having a different angle QPR how can you prove it is 45 degrees?

Shouldn't this question be find the points P, Q and R where angle QPR is 45 degrees?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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I am sorry

I will try to find out the original question from whom I got it once this formative assesment is over

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,897

Hi;

I have a solution that finds the points you need. My solution shows that PRB is a right angle which is not shown in your drawing or given.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

**Online**

You seem correct.

How about the proof?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,009

hi Agnishom and bobbym

Here's two new diagrams.

I constructed the square ABCD.

I put point G somewhere on AD

I made an isosceles triangle FGC so that FG = FC

I found the centre of the square and rotated point F by 270 degrees around this point to fix H.

That makes BH = FG = FC

I constructed a line, perpendicular to GF at G and bisected the angle to make FGJ = 45.

As you can see in the first diagram J and H are different points.

But, this method of construction allows me to move G along the line. My plan was to find the position where J and H coincide.

Because of the size of the points, it's hard to be exact about this (Euclid had dimensionless points but they cost extra )

So I measured GHB and GJB and tried to make them equal. The second diagram shows my best attempt.

So the 90 case seems to be a special case and the only one where FGH = 45.

(Not counting G at A or B)

Your original diagram does make it look like there's a right angle there, but, in geometry, it is dangerous assuming things just because they look like it.

Later I'll have a go at the proof when I assume the angle is 90.

Bob

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: 7,009

I think you can do this yourself.

Hint: draw the line FK so that K is on GH and FK is perpendicular to GH.

Bob

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Good Heavens! What a confusion!

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,897

You seem correct.

How about the proof?

Actually, the diagram is the proof. A proof without words. It uses coordinate geometry and trigonometry to get every side and every angle. To have that drawing one must find that angle PRB is a right angle which I said earlier.

The diagram does not mention the right angle so finding the solution, where P, Q and R is all that is necessary. To prove that angle is 45 degrees was not possible without the given of angle PRB as 90 degrees.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

**Online**

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