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.

]]>Good Heavens! What a confusion!]]>

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

Bob

]]>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

]]>How about the proof?]]>

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.

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

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?

]]>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

]]>Hhhmm. as given, that diagram is impossible.

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

Bob

]]>Are you sure it is DR?

]]>PQ = QC = DR

Prove that: angle QPR = 45 degrees

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

]]>