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. FLH 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 EMC From I draw IK perpendicular to AC; then (Eucl. vi. 3.) DA: AC:: DF: FC: DG : (GA=) 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 Ꮐ 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 H C 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. 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 :: DE GH'. 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. H G Join AF; and let HI cut AB in K. Since IG is parallel to AD, it is perpendicular to AB; .. the angle GKO is a right angle, and equal to AFO; and the vertical angles at O are equal, and GOOA; .. the triBut angles GKO, OFA are equal, and OF=OK. since OB-BE, .. (Eucl. vi. 2.) OK-KI; and .. OF =KI, and OG=KH; .. FG=IH. (64.) If two sides of a trapezium inscribed in a circle be produced, and from the same point in one side produced a line be drawn parallel to the other, intersecting the adjacent side of the trapezium, and a second line to the extremity of that other intersecting the circumference; the line joining the two points of intersection, will pass through the same point. Let the two sides AD, BC, of the trapezium ABCD inscribed in the circle ABC, be produced, and let them meet in E; and from any point in AD produced, draw FH parallel to BE, meeting the side DC in H; and join FB, meeting the circumference in G; the line joining G, H will always pass through the same point. F F H Let GH produced meet the circle in I. Join AI, DG. The angle GDH=GBC in the same segment, and .. is equal to the alternate angle GFH; whence a circle may be described through the points G, H, D, F; and .. the angle DGH=DFH =DEB. But the angle DEB being always the same, DGI, and .. DAI, and also the arc DI will be invariable, and D being a fixed point, I must be also; i. e. GH will always pass through I. FF (65.) If the diagonals of a quadrilateral figure inscribed in a circle cut each other at right angles, the rectangles contained by the opposite sides are together double of the quadrilateral figure. Let ABCD be a quadrilateral figure inscribed in a circle, whose diagonals AC, BD cut each other at right angles in E; the rectangles contained by AB, CD, and AD, BC are together double of the figure. B D E For (Eucl. vi. D.) the rectangles contained by AB, CD, and AD, BC, are together equal to the rectangle AC, BD, i. e. to the rectangles contained by AC, BE and AC, ED. But the rectangle contained by AC, BE is double of the triangle ABC, and the rectangle contained by AC, ED, is double of ADC; hence the rectangles contained by AB, CD and AD, BC are together double of ABCD. (66.) If a rectangular parallelogram be inscribed in a right-angled triangle, and they have the right angle common; the rectangle contained by the segments of the hypothenuse is equal to the sum of the rectangles contained by the segments of the sides about the right angle. Let ABC be a right-angled triangle, in which the rectangular parallelogram DBEF is inscribed, having one of its angles at B; the rectangle AF, FC is equal to the rectangles AD, DB and BE, Draw EG perpendicular to FC. ADF, EFG being similar, D B E G EC together. The triangles AD AF FG (EF=) BD, |