This proof is shorter than the usual one, since it is not necessary to prove that AC, CD are in one right line. In a similar way the Proposition may be proved by taking any of the eight figures formed by turning the squares in all possible directions. Another simplification of the proof would be got by considering that the point A is such that one of the △s CAG, BAK can be turned round it in its own plane until it coincides with the other; and hence that they are congruent. Exercises. 1. The square on AC is equal to the rectangle AB.AO, and the square on BC = AB.BO. 2. The square on CO = AO.OB. 3. AC2 − BC2 = AO2 − BO2. 4. Find a line whose square shall be equal to the sum of two given squares. 5. Given the base of a triangle and the difference of the squares of its sides, the locus of its vertex is a right line perpendicular to the base. 6. The transverse lines BK, CG are perpendicular to each other. 7. If EG be joined, its square is equal to AC2 + 4BC2. 8. The square described on the sum of the sides of a right-angled triangle exceeds the square on the hypotenuse by four times the area of the triangle (see fig., xlvi., Ex. 3). More generally, if the vertical angle of a triangle be equal to the angle of a regular polygon of n sides, then the regular polygon of n sides, described on a line equal to the sum of its sides, exceeds the area of the regular polygon of n sides described on the base by n times the area of the triangle. 9. If AC and BK intersect in P, and through P a line be drawn parallel to BC, meeting AB in Q; then CP is equal to PQ. 10. Each of the triangles AGK and BEF, formed by joining adjacent corners of the squares, is equal to the right-angled triangle ABC. 11. Find a line whose square shall be equal to the difference of the squares on two lines. 12. The square on the difference of the sides AC, CB is less than the square on the hypotenuse by four times the area of the triangle. 13. If AE be joined, the lines AE, BK, CL, are concurrent. 14. In an equilateral triangle, three times the square on any side is equal to four times the square on the perpendicular to it from the opposite vertex. 15. On BE, a part of the side BC of a square ABCD, is described the square BEFG, having its side BG in the continuation of AB; it is required to divide the figure AGFECD into three parts which will form a square. 16. Four times the sum of the squares on the medians which bisect the sides of a right-angled triangle is equal to five times the square on the hypotenuse. 17. If perpendiculars be let fall on the sides of a polygon from any point, dividing each side into two segments, the sum of the squares on one set of alternate segments is equal to the sum of the squares on the remaining set.