parallel to BC, the angle FHE is equal to HEB. Also since the circles are equal, the arc HB is equal to the arc FE (ii. 1. Cor. 1.), .. the angle HEB is equal to FCE, .. FCE = .. FCE = FHE, and HC is a parallelogram; whence HF-EC= CF. Also since the rectangle CF, FA is equal to the rectangle HF, FG, and HF=CF, .. FA=FG. COR. Hence if any number of lines be drawn parallel to BE, and terminated by the two circumferences, each of them will be equal to BE. (58.) If two diagonals of an equilateral and equiangular pentagon be drawn to cut one another, the greater segments will be equal to the side of the pentagon; and the diagonals cut one another in extreme and mean ratio. E Let ABDCE be an equilateral and equiangular pentagon; draw the diagonals ED, BC cutting each other in F; EF and FB will be each equal to a side of the pentagon; and ED, BC are cut at F, in extreme and mean ratio. About the pentagon describe a circle. And since AB=CE, the arcs AB, CE are equal; .. AE is parallel to BC. For the same reason, AB is parallel to EF;.. the figure ABFE is a parallelogram; whence AB=FE, and AE= FB; but AB= AE, .. EF= FB, and each is equal to a side of the pentagon. Also the angle DCF=CDF=DEC, .. the triangles DCF, DEC are similar, and ED CD :: CD : DF, or ED: EF:: EF: FD; .. ED is cut in extreme and mean ratio. The same may also be proved of BC. (59.) If the sides of a triangle inscribed in the segment of a circle be produced to meet lines drawn from the extremities of the buse, forming with it angles equal to the angle in the segment; the rectangle contained by these lines will be equal to the square described on the base. Let the sides AB, CB of the triangle ABC, inscribed in the segment ABC, be produced to meet CE, AD, which make with AC, angles equal to the angle ABC in the segment; the rectangle AD, CE is equal to the square of AC. A Since the angle ABC=DAC, and the angle at C is common to the triangles ABC, ADC, the triangles are similar. In the same manner it may be shewn that ABC, AEC are similar; and .. ADC, AEC are also similar; whence AD AC: AC : CE, and the rectangle AD, CE is equal to the square of AC. (60.) If two triangles (one of them right angled) have the same base and altitude, and the hypothenuse intersect a line which is drawn bisecting the right angle; a line passing through this point of intersection parallel to the base, and terminated by the sides of the other triangle, shall be a side of the square inscribed within it. G B FL H Let ADC be a right-angled triangle, and ABC on the same base, have its altitude BEAD; and let the hypothenuse DC, meet AF which bisects the angle DAC in F; through which draw GH parallel to AC; IH will be the side of a square inscribed in the triangle ABC. K E M C From I draw IK perpendicular to AC; then (Eucl. vi. 3.) DA: AC: DF: FC: DG (GA=) IK, also AC BE:: IH: (BL=) DG, : .. ex æquo DA: BE :: IH: IK; But DA=BE, .. IH=IK; and if HM be drawn perpendicular to AC, IM is a parallelogram, whose sides are equal; and the angles at K and M being right angles (Eucl. i. 46. Cor.) it is a square. (61.) If on the side of a rectangular parallelogram as a diameter, a semicircle be described, and from any point in the circumference lines be drawn through its extremities to meet the opposite side produced; the altitude of the parallelogram will be a mean proportional between the segments cut off. On AB, the side of the rectangular parallelogram ABCD, let a semicircle AEB be described; and from any point E, draw EA, EB, and produce G E them to meet CD produced; AD will be a mean proportional between GD and CF. Since DG is parallel to BA, the angle DGA is equal to BAE, and the angles at D and E are right angles, .. the triangles BAE, DGA are equiangular. In the same manner it may be shewn that FCB is equiangular to BAE, and .. to DGA; whence GD: DA :: (CB=) DA: CF. (62.) If on the diameter of a semicircle a rectangular parallelogram be described, whose altitude is equal to the chord of half the semicircle, and lines drawn from any point in the circumference to the extremities of the base intersect the diameter; the squares of the distances of each point of section from the farthest extremity of the diameter will be together equal to the square of the diameter. A B F G H Let ABC be a semicircle, on the diameter of which describe the rectangular parallelogram AE, whose side AD is equal to AB a chord of half ABC; and from any point F in the semicircle draw FD, FE cutting the diameter in G and H; the squares of AH and CG are together equal to the square of AC. D K D Draw the perpendicular FK; the triangles DGA, DFK being similar, DA: AG :: FK : KD, and ECH, FKE being similar, .. DA: AG× CH :: FK': (KE × KD=) IF Now the square of DE is double of the square of DA, ..the square of GH is double of the rectangle AG, CH. But the square of AH is equal to the squares of AG, GH and twice the rectangle AG, GH, i. e. to the square of AG and twice the rectangle AG, GC; ;.. the squares of AH and GC are together equal to the squares of AG, GC, and twice the rectangle AG, GC, i. e. to the square of AC, (Eucl. ii. 4.). COR. The square of the part of the diameter intercepted between the two lines drawn from the point in the semicircle is double of the rectangle contained by the two extreme segments. (63.) If on the radius drawn from the point of contact of a circle and its circumscribed square, another circle be described; and from any point in the outer circumference a line be drawn through its centre to the inner circumference, and through the same point another line be drawn parallel to the common tangent to the circles, and terminated by the side of the square and its diagonal; these two lines are equal. Let O be the centre of the circle, circumscribed by a square, whose diagonal is DE. On 40 describe a circle AOF; and from any point F draw a line FOG; and through G draw HI parallel to AD; FG is equal to HI. Π D B E G K A Join AF; and let HI cut AB in K. Since IG is |