Solutions for Session 6, Part B

See solutions for Problems: B1 | B2 | B3 | B4

 Problem B1 By construction, all four triangles are right triangles. The longer leg of each triangle has length b (since it equals the length of the side of the larger inscribed square), and the shorter leg of each triangle has length a (since it equals the length of the side of the smaller inscribed square). So all four triangles have two congruent sides as well as the angle between the two sides. So, by SAS congruence, the four triangles are indeed congruent to the original.

 Problem B2 The sides of the piece in the center are all equal to c, since they are the longest sides of four triangles that are congruent to the original triangle (see the argument of Problem B1). Now we have to show that the angles are each 90°. Notice that three angles line up to make a straight line, or 180°: the small and the middle angles from the original right triangle (marked with one and two lines, respectively), and the one from the quadrilateral that lies between the two (see picture). Since the three angles of a triangle also add up to 180°, the angle in the quadrilateral must be the same as the largest angle in the triangle -- a right angle. This is true for all four angles in the shape, and so we know that we have a quadrilateral with four congruent sides and four right angles, so it must be a square.

Problem B3

Here is an interactive illustration that describes the reasoning behind "Behold!"