Theorem 3.4: (Pythagoras) Let   f   d,   and   r   be the three sides of a triangle.

The triangle is a right triangle if and only if

a2 + f2 = r2

 

Proof: If the lines determined by   OF   and   FA   are horizontal or vertical, then the result follows from the distance formula, so we can assume that the line determined by   FA   is not vertical or horizontal, in which case it will have an equation of the form   y = mx + b   by Theorem 1.2,   (m not 0).

Assume the triangle is a right triangle, with   r   the hypoteneuse. Let   O   and   A   be the end points of the hypoteneuse, and let   F   be the point at the right angle. Consider the circle centered at   O   with radius   r.

Let   O = (xo, yo),   and let   A = (x1, y1).   By Theorem 3.3, the line may intersect the circle at another point which we will call   B = (x2, y2).   Then the equation of the line will be   y = mx + b,   where   b   is the   y-intercept pictured above, and   m   is the slope computed using the points   (x1, y1)   and   (x2, y2).   Since   OF   and   AF   are perpendicular,   F   is the foot of   (xo, yo),   the center of the circle in the line   y = mx + b,   so by Theorem 1.10 its coordinates are

By Theorem 3.3,

so by Theorem 1.5,

But, by Theorem 1.11, the perpendicular distance from the point   O   to the line   AF   is

so

Square both sides

a2 = r2 - f 2

and add f2 to both sides.

a2 + f 2 = r 2

For the converse, assume that   a2 + f2 = r2

a2 = (x2 - x3 )2 + (y2 - y3 )2

f2 = (x1 - x3 )2 + (y1 - y3 )2

r2 = (x2 - x1 )2 + (y2 - y1 )2

so, if

a2 + f2 = r2

then

(x2 - x3 )2 + (y2 - y3 )2 + (x1 - x3 )2 + (y1 - y3 )2 = (x2 - x1 )2 + (y2 - y1 )2

x22 - 2x2x3 + x32 + y22 - 2y2y3 + y32 + x12 - 2x1x3 + x32 + y12 - 2y1y3 + y32 = x12 - 2x1x2 + x22 + y12 - 2y1y2 + y22

Subtract   x22 + x12 + y22 + y12   from both sides and combine the   x3   and   y3   terms.

- 2x2x3 + 2x32 - 2y2y3 + 2y32 - 2x1x3 - 2y1y3 =- 2x1x2 - 2y1y2

Divide both sides by   2.

- x2x3 + x32 - y2y3 + y32 - x1x3 - y1y3 = -x1x2 - y1y2

Transpose

x1x2- x2x3- x1x3 + x32+ y1y2- y1y3 - y2y3+ y32 = 0

These factor

(x1 - x3)(x2 - x3) + (y1 - y3)(y2 - y3) = 0

Transpose

(x1 - x3)(x2 - x3) = -(y1 - y3)(y2 - y3)

Divide

and   OF   is perpendicular to   AF.1 2

 

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