# Point on circle about square

1. Jun 22, 2009

### veljko_c

Let on side BC of square ABCD set point E, BE:EC=1:2. On side DC point F
(D-C-F), CF:DF=1:3.
Prove that point M = intersection AE and BF, is on circle about ABCD.

Thnx

2. Jun 22, 2009

### HallsofIvy

Staff Emeritus
Here's one way, not necessarly the best but just because I tend to be "analytic". Set up a coordinate system in which the origin is at the center of the square, side AB is x= -1, side CD is x= 1, side BC is y= -1, and side AD is y= -1.

B is (-1, -1) and C is (1, -1) so E is (x,-1) with 1- x= 2(x- (-1)) so 1- x= 2x+ 2 or 3x= -1 so x= -1/3. E is (-1/3, -1) and the line through A and E, the line through (-1, 1) and (-1/3,-1), is y= -3(x+1)+ 1= -3x- 2.

D is (1, 1) so F is (1, y) with y+1= 3(y- 1)= 3(y- 1) so y+ 1= 3y- 3 or 2y= 4 so y= 2. F is (1, 2) and the line through B and F, the line through (-1, -1) and (1, 2) is y= (3/2)(x+1)-1= (3/2)x+ 1/2.

M is the point at which those two lines intersect so you can find it by solving the two equation simultaneously. Then find the distance from that point to the center of the square, (0,0), and show that is the same as the distance from (0,0) to each of the vertices of the square.

3. Jun 22, 2009

### tiny-tim

Welcome to PF!

Hi veljko_c! Welcome to PF!
I'm confused … the way I've drawn it, they intersect inside the square.

4. Jun 24, 2009

### chicopee

I am a little bit "befuddled" about the sentence "point on a circle about square", however I am assuming that you are referring about four corners of a square on a circle . The proper sentence should be " circumscribing the square on the circle" that way there is no confusion.

Nonetheless, letting AB=BC=CD=DA =a; E=a/3 from point B; F=a/4 from point C. Point M, by visual examination, is inside the square and the circle and is not at the center of the circle.

5. Jun 24, 2009

### tiny-tim

Oh I get it now …

M is on the circle that fits inside the square …​

so define x and y coordinates for A B C and D, find the gradients of AE and BF, and write an equation y = (gradient)x + constant for both, and eliminate y, and show that x2 + y2 = AB2/4

6. Jun 25, 2009

### chicopee

Tiny Tim, are you saying the the square inscribes the circle, another word, the sides of the square are tangent to the circle?

7. Jun 30, 2009

### chicopee

I just think that the original question was badly worded.
Good going on the solution Tiny-Tim

8. Jun 30, 2009

### belliott4488

We need the OP to clarify the problem, but I took the parenthetical expression "(D-C-F)" to mean that the point F is on the extended line CF, i.e. it is not on the side of the square but is outside of it. If you draw it this way, then the two lines intersect outside the square.

I believe this is how HallsOfIvy interpreted it as well, although I would have put point F at (1,-2), taking the ordering D-C-F to mean that F must be below the point C.