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- bobbym
- 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.
- Agnishom
- 2012-11-29 14:55:57
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
- 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 
)
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
- Agnishom
- 2012-11-28 19:23:10
You seem correct.
How about the proof?
- bobbym
- 2012-11-28 14:22:52
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.
- Agnishom
- 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
- bobbym
- 2012-11-28 10:13:13
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?
- 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.
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
- Agnishom
- 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.
Bob
- bobbym
- 2012-11-27 17:54:39
Hi Agnishom;
Are you sure it is DR?
- Agnishom
- 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