The construction of the following figure is evident. It is easily shown that the four quadrilaterals ABMX, XNCA, SBCP, and SRQP are congruent.
∴ ABMXNCA equals SBCPQRS but is not congruent to it, the congruent quadrilaterals being differently arranged.
Subtract the congruent triangles MXN, ABC, RAQ, and the proposition is proved.[80]
The following is an interesting proof of the proposition:
Let ABC be the original triangle, with AB < BC. Turn the triangle about B, through 90°, until it comes into the position A'BC'. Then because it has been turned through 90°, C'A'P will be perpendicular to AC. Then
1/2(AB)2 = ⧍ABA',
and 1/2(BC')2 = ⧍BC'C,
because BC = BC'.
∴ 1/2((AB)2 + (BC)2) = ⧍ABA' + ⧍BC'C.
∴ 1/2((AB)2 + (BC)2)
= ⧍AC'A' + ⧍A'C'C
(For ⧍ABA' + ⧍BC'A' + ⧍A'C'C is the second member of both equations.)
= 1/2A'C' · AP
+ 1/2A'C' · PC
= 1/2A'C' · AC
= 1/2(AC)2.
∴ (AB)2 + (BC)2 = (AC)2.