How can you prove the pythagorean theorem and use it to solve problems?

Answer 1

There are two ways to prove the theorem: one graphically (visually), one algebraically:

They both start in the same way (this is going to take some explaining as I can't draw a diagram very easily - try and draw the diagram on a piece of paper from the instructions):

Take a square ABCD of side length a + b.

Mark E at distance 'a' from A along AB (which is distance 'b' from B along AB)

Mark F at distance 'a' from B along BC (which is distance 'b' from C along BC)

Mark G at distance 'a' from C along CD (which is distance 'b' from D along CD)

Mark H at distance 'a' from D along DA (which is distance 'b' from A along DA)

Now draw in: EF, FG, GH, HE.

This forms four right angled triangles: AEH, BFE, CGF, DHG

And a quadrilateral EFGH.

Using the above diagram we can show:

The four right angled triangles are congruent using Side-Angle-Side on the sides with lengths 'a' and 'b' and the enclosed right angle. Thus the hypotenuses of the four triangles are the same and can be labelled distance 'c'.

Next, the sides of the quadrilateral EFGH are the hypotenuses of the triangles which are all distance 'c', so EFGH has four sides of equal length.

Now consider angle HEF:

Since the angles in a triangle sum to 180°

AEH + EHA + HAE = 180°, thus AEH + EHA = 180° - HAE = 180° - 90° = 90°

Since triangles AEH and BFE are congruent, angle EHA = angle FEB

Thus the sum of angles AEH + FEB = 90°

E is along the line AB so angle AEB is 180°.

Thus the sum of angles AEH + FEB + HEF = 180° but AEH + FEB = 90°, thus HEF = 180° - 90° = 90°.

Similarly for all the angles of the quadrilateral EFGH, thus EFGH is a square of side length 'c'.

(Starting with four congruent triangles and arranging them in the big square of the diagram is easier to start the proof, but harder to explain.)

Thus the diagram shows four congruent triangles of side lengths a, b, c arranged around a square of side length c forming a square of side length a + b.

Graphical proof:

Take triangle AEH and translate it so that HE coincides with BF; the new position of A is called A' and the rectangle formed is A'FCB

Take triangle EBF and translate it so that E coincides with A; where B ends up on AB call B' and the new position of F call F'

Finally take triangle DHB and translate it to that HB coincides with AF; the new position of D is called D' to form a rectangle AB'F'D'.

Length B'F' is 'a' so the distance from CD of F' is 'b'.

Length D'F' is 'b' so the distance from BC of F' is 'a'.

But GA' is length 'b' and FA' is length 'a' so F' and A' coincide

Thus the square ABCD is now divided up into two rectangles (each made up of two of the original four triangles) and two squares: B'BFA' (of side length 'a') and D'A'GD (of side length 'b')

As the area of the triangles has not changed, the area of the square EFGH (of side length 'c') must equal the sum of the areas of these two small squares.

Thus c² = a² + b²

[This is the image above]

Algebraic proof:

Consider the area of the big square. It can be calculated as the square of the side length or as the sum of the areas of the four triangles and the smaller square:

area = (a + b)² = 4 x ½ ab + c²

→ a² + 2ab + b² = 2ab + c²

→ a² + b² = c²

But as 'c is the length of the hypotenuse of one of the triangles, and 'b' & 'c' are the lengths of the other two sides, this proves Pythagoras' Theorem.

One example where Pythagoras' Theorem is used in reverse: if a² + b² = c² then the angle opposite 'c' is a right angle. Making a triangle with sides 3, 4, 5 ensures a right angle is opposite the 5 side and this is used in building work to ensure construction is square.