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Pythagorean Theorem Proof using Circles

This is outcome of boredom in managment classroom :)

Pythagoras Theorem proof using properties of Circle and tangent

Consider triangle ABC, right angled at B. To prove

AB^2+BC^2=AC^2

Construction:

Draw circle X with AB as diameter. This circle cuts AC at O. Construct another circle Y with BC as diameter. Circle cuts AC at P.

From properties of Tangent and circle
Now AB is tangent to circle Y at B. Hence

AB^2 = (AP)(AC) ……(i)
Similarly, BC is tangent to circle X at B.

BC^2 = (OC)(AC) …(ii)

Adding (i) and (ii)

AB^2 + BC^2 = (AP)(AC) + (OC)(AC) = (AC)(AP + OC)

Now since AB is diameter to Circle X, Angle AOB is right angled. Similarly P is on circle Y with BC diameter. Hence angle BPC is right angled. Thus BP as well as BO is perpendicular to line AC; which is not possible unless they are same point. Hence OP = 0.

or

AP + OC = AC   …(iii)

Hence AB^2 + BC^2 = AC^2

Somebody has arrived at this proff earlier :(

http://mathforum.org/library/drmath/view/55219.html

so what?
maza raaste mein hai …manzil mein nahi. :)

What properties are being used? (Which tangent?)

AB is tangent to bigger circle.
section 20.8 INTERSECTING SECANTS AND TANGENTS OF A CIRCLE of http://www.nos.org/Secmathcour/eng/ch-20.pdf