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## #1 2012-11-27 13:07:33

Agnishom
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### Prove this

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'
'The whole person changes, why can't a habit?' -Alokananda

## #2 2012-11-27 17:54:39

bobbym

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### Re: Prove this

Hi Agnishom;

Are you sure it is DR?

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #3 2012-11-27 19:03:36

bob bundy
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### Re: Prove this

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

## #4 2012-11-28 00:08:15

Agnishom
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### Re: Prove this

'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'
'The whole person changes, why can't a habit?' -Alokananda

## #5 2012-11-28 01:03:03

bob bundy
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### Re: Prove this

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

## #6 2012-11-28 10:13:13

bobbym

Online

### Re: Prove this

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.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #7 2012-11-28 14:19:06

Agnishom
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### Re: Prove this

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'
'The whole person changes, why can't a habit?' -Alokananda

## #8 2012-11-28 14:22:52

bobbym

Online

### Re: Prove this

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.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #9 2012-11-28 19:23:10

Agnishom
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### Re: Prove this

You seem correct.

'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'
'The whole person changes, why can't a habit?' -Alokananda

## #10 2012-11-28 20:45:16

bob bundy
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### Re: Prove this

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

## #11 2012-11-28 22:59:44

bob bundy
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### Re: Prove this

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

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

## #12 2012-11-29 14:55:57

Agnishom
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### Re: Prove this

Good Heavens! What a confusion!

'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'
'The whole person changes, why can't a habit?' -Alokananda

## #13 2012-11-29 19:28:40

bobbym

Online

### Re: Prove this

You seem correct.