whence, AC2+DB2 = AD2 + CB2 + 2AB2+2AB × BE+2AB × AF. Now (Eucl. ii. 1.), AB × FE=AB × FA+AB× AB+AB × BE, ··. AC2 + DB2 = AD2 + CB2+2 AB × DC. (36.) The squares of the diagonals of a trapezium are together double the squares of the two lines joining the bisections of the opposite sides. Let ABCD be a trapezium, whose sides are bisected in E, F, G, H. Join EG, FH; and draw the diagonals AC, BD. The squares of AC, BD are together double of the squares of EG, FH. H E B Join EF, FG, GH, HE. Then (iv. 14.) EFGH is a parallelogram, and BD is double of EH; .. BD2 = 4EH2 = 2 EH2 + 2 FG3, and for the same reason AC2 = 2 EF2 + 2HG, (37.) The squares of the diagonals of a trapezium are together less than the squares of the four sides, by four times the square of the line joining the points of bisection of the diagonals. Let ABCD be a trapezium whose diagonals AC, BD are bisected in E, F; join EF; the squares of AC, BD are less than the squares of the four sides by four times the square of EF. Since BE bisects AC the base of the triangle ABC, AB+BC2 AE2 + 2 EB; and for a similar reason, AD2+ DC2=2AE2+2 ED2; F E D .. AB2+BC2+CD2 + DA2 = 4 AE2 + 2 EB' + 2 ED' B (38.) In any trapezium, if two opposite sides be bisected; the sum of the squares of the two other sides, together with the squares of the diagonals, is equal to the sum of the squares of the bisected sides together with four times the square of the line joining those points of bisection. Let AB, DC, two opposite sides of the trapezium ABCD, be bisected in E, and F; join EF; and draw the diagonals AC, BD. The squares of AD, BC, AC, BD are equal to the squares of AB, DC, and four times the square of EF. Join AF, BF. Since AF bisects DC the base of the triangle ADC, AD2 + AC2 = 2 DF2 + 2 FA2; and in the same manner, BCBD = 2 DF + 2 FB2; whence AD2+BC2+AC2+BD'≈4 DF2+2 FA2+2 FB2 = = DC2 + 2 FA2 + 2 FB2 = DC2+4 AE2 + 4 EF2 (39.) If squares be described on the sides of a rightangled triangle; each of the lines joining the acute angles and the opposite angle of the square, will cut off from the triangle an obtuse-angled triangle, which will be equal to that cut off from the square by a line drawn from the intersection with the side to that angle of the square which is opposite to it. From the angles B, C of the right-angled triangle BAC, let lines BG, CD be drawn to the angles of the squares described upon the sides, and from the intersections H and I let HE, IF D H B A be drawn to the opposite angles of the squares; the triangle BIC=AIF, and CHB = AHE. G Join AG, AD. Then (Eucl. i. 37.) the triangle AFI = AIG; to each of which add ABI, .. the triangle BIF = BAG = BCA (Eucl. i. 37.) From each of these equals take away the triangle BIA, and BIC= AIF. In the same manner it may be shewn that CHB = AHE. (40.) If squares be described on the two sides of a right-angled triangle; the lines joining each of the acute angles of the triangle and the opposite angle of the square will meet the perpendicular drawn from the right angle upon the hypothenuse, in the same point. Let BE, CF be squares described on the sides BA, AC containing the right angle. Join DC, BG; they L จ intersect AL, which is perpendicular to BC, in the same point O. Produce DE, GF, to meet in H. Join HA, HB, HC. Let BH, CH respectively meet DC, BG in I and K. Since EH- AF-AC, and EA= AB, and the angles HEA, BAC are right angles, the triangles HEA, BAC are equal, and the angle EHA = BCA =BAL, i. e. since EH and BA are parallel, HAL is a straight line, or LA produced passes through H, and HL is perpendicular to BC. Again, since AC=CG, AH=BC, and the angle HAC = BCG, :. the triangles HAC, BCG are equal; .. the angle CBK=CHL; but BCK HCL; .. BKC=HLC, i, e. is a right angle, and BK is perpendicular to HC. In the same manner it may be shewn that CI is perpendicular to BH. Hence .. HL, CI, BK are perpendicular to the sides of the triangle HBC, and .. they intersect each other in the same point. = (41.) If squares be described on the three sides of a right-angled triangle, and the extremities of the ad T jacent sides be joined; the triangles so formed are equal to the given triangle and to each other. On the sides of the rightangled triangle ABC let squares be described, and join GH, FD, IE. The triangles AGH, BFD, ECI are equal to ABC, and to each other. It is evident that AGH= ABC. Produce FB, and from D draw DS perpendicular to F H B D C R E it. Since ABS and CBD are right angles, .. the angles ABC, SBD are equal; and BAC, BSD are also right angles, and BC= BD, .. DS=AC. And the triangles ABC, FBD being upon equal bases AB, FB are as their altitudes AC, DS (Eucl. vi. 1.); and .. are equal. In the same manner if IC be produced, and ER drawn perpendicular to it, it may be shewn that ER is equal to AB, and the triangle ECI to ABC. And since each of the triangles is equal to ABC, they are equal to one another. (42.) If the sides of the square described upon the hypothenuse of a right-angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the legs; they will cut off triangles equiangular and equal to the given triangle. Let DB, EC, the sides of the square described on BC the hypothenuse of the right-angled triangle ABC, be produced to meet the sides of the squares described |