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Topic review (newest first)

2012-11-29 19:28:40

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.

2012-11-29 14:55:57

roll dizzy
Good Heavens! What a confusion!

bob bundy
2012-11-28 22:59:44

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 bundy
2012-11-28 20:45:16

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 smile  )

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.


2012-11-28 19:23:10

You seem correct.
How about the proof?

2012-11-28 14:22:52


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.

2012-11-28 14:19:06

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

2012-11-28 10:13:13


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?

bob bundy
2012-11-28 01:03:03

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.


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.


2012-11-28 00:08:15

How about it being BR?

bob bundy
2012-11-27 19:03:36

hi Agnishom,

Hhhmm.  as given, that diagram is impossible.

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


2012-11-27 17:54:39

Hi Agnishom;

Are you sure it is DR?

2012-11-27 13:07:33

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

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