*Hubert Shutrick*

This proof is perhaps the easiest to understand for a person who knows very little mathematics and could well be used for schoolchildren.

**Translation** A copy of a diagram can be made by first drawing it on (transparent) tracing paper laid over the diagram and then sliding it to the new position, If this is done without turning the tracing paper, the new diagram is a **translation** of the original, lines in the copy are parallel with the corresponding ones in the original and areas are preserved.

**Pythagoras Theorem** In the diagram, ABC is the triangle with the right angle at C and ADEB is the square on the hypotenuse AB . Pythagoras theorem claims the the area of ADEB is the sum of the areas of the squares on the other two sides.

**Construction** Divide the square on BC into four areas with the line segment that extends EB and the one through C parallel with AB , the translation of AB . If C' is the point on AB such that the angle BC'C is 90° , then the triangle A'B'C' is the translation of ABC obtained by sliding C to C' . The triangle C'B''C'' is obtained by first rotating the image of ABC 90° clockwise about A and then translating with A going to C' . Finally, the square on AC is translated to C'A'FC'' , where F is on B''C'' .

**Proof** If we ignore the segments A'B' and C'B'' , the part of ADEB outside of C'A'FC'' is divided into four areas that are clearly translations of the areas in the square on BC .

**Reference** I found this diagram in some lecture notes of Professor Hanner about forty years ago. He might have been studying dissections of the plane similar to those in G. N. Frederickson, Dissections: Plane & Fancy, Cambridge University Press, 1997.