Proof B of the 12th Century Hindu mathematician Bhaskara.
Given the figure, 4 right triangles and a rectangle combined to make a larger rectangle, it
can be proven that the larger rectangle is more specifically a square:
Using the Side Angle Side Theorem and the sides a and b of the right triangle
along with the right angles of the triangle, it can be shown that all four triangles are
congruent. Since we know the four triangles are congruent we also know that all four
hypotenuses are equal in length. With the knowledge that all the angles in a single
triangle add up to 180° it can also be known that angle a and angle b combine to equal
90°. Since the larger square’s angles are composed of one angle a and one angle b it can
be known that the angles of the rectangle are 90°. A rectangle composed of equal side
lengths and 90° angles is known as a square.
With the given information we can also prove that the inner rectangle is a square as well:
All the sides of the inner rectangle are equal to the length a minus b giving the
rectangle equal sides lengths. We know that the angles of the inner rectangle are equal to
90° because if a line is intersected with another line both angles created by the lines add
up to equal 180°. Since one angle of the intersected line equals 90°, the other angle must
equal 90° as well. Now that we know that the inner rectangle is composed of all equal
side lengths and 90° angles we know that the inner rectangle is a square as well.
The Pythagorean Theorem can be proved by equating the area of the large square (c2) to
the total areas of the four right triangles 4(½ab) and the inner square (a-b) 2 :
4( ½ab ) + ( a-b ) 2 = c2
By multiplying four times (½ba) and foiling (a-b) 2 we get
2ab + a2 – 2ab + b2 = c2
The negative and positive 2ab cancel out and we are left with
a2 + b2 = c2
Also known a the Pythagorean Theorem