Hi Agnishom;

Are you sure it is DR?

How about it being BR?

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?

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.

Uploaded Images

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

Uploaded Images

Good Heavens! What a confusion!