So, would you get the bonus marks? :)

In triangle ABC, B = 90 degrees. A square is drawn on the hypotenuse AC and P is the centre of the square. Prove that angle PBC = 45 degrees.

This is how far I got:

/\
/ \
/ \ THis, of course, is supposed to represent a square,
/ \ sorry for the lame-o drawing
A / \
|\ p /
| \ /
| \ /
| \ /
|____\/
B C

let M be midpoint of AC
M = 1/2A + 1/2C
= (c/2, a/2)

If AM = BM = CM then triangle AMB = triangle BMC
_____________
|AM| = |c^2/4 +a^2/4
\|
_____________
|BM| = |c^2/4 +a^2/4
\|
_____________
|CM| = |c^2/4 =a^2/4
\|
therefore AM=BM=CM

but then I realized that this was faulty logic, because this only applies if it's an isosceles triangle. My teacher was trying to get it to be 1 over root 2 with trig identities, but wasn't able to get it. I'm just curious to see if it actually works out or if there's a problem with the question.

Thanks,
Ange
(what's the circumfrence of an igloo?)

...an eskimo pi...

If the story is true, you should probably seek a different school.

>So, would you get the bonus marks? :)
>
>In triangle ABC, B = 90 degrees. A square is drawn on the
>hypotenuse AC and P is the centre of the square. Prove that
>angle PBC = 45 degrees.
>
>This is how far I got:
>
> /\
> / \
> / \ THis, of course, is supposed to represent a
>square,
> / \ sorry for the lame-o drawing
>A / \
> |\ p /
> | \ /
> | \ /
> | \ /
> |____\/
>B C
>
>
>
>
>
>
>let M be midpoint of AC
>M = 1/2A + 1/2C
> = (c/2, a/2)
>
>If AM = BM = CM then triangle AMB = triangle BMC

No, this is incorrect. In any right triangle, AM = BM = CM, where M is the midpoint of hypotenuse.

>I'm just curious to see if it
>actually works out or if there's a problem with the
>question.

Since APC is also a right triangle,

AM = BM = PM.

So that A, B, C, P all lie on a circle with center M. AC is its diameter. P lies on the perpendicular to chord through its center.

The rest you must be able to do by yourself.

I just see your question and since is an old one probably you got the answer

Anyway I believe the following is what you are looking for

As you already found BM= AM= CM

What indeed means the point M lays always in a circumference centered on B and radius L if we call 2L the length of hypotenuse AC. That’s also means triangle BMC is isosceles and the sum of the angles u and w is 90

w + u = π / 2 ; u = π / 2 – w

Then coordinates for P are :

Xp = L cos w + L cos(π / 2 – w) = L< cos w + sin w >
Yp = L sin w + L sin (π / 2 – w) = L< sin w + cos w >

So Yp/ X p the slope of line OP is always equal one what means line OP forms an angle π / 4 to each of the axis.

This an analytical solution of course there are other more elegant using simple geometry.