D and E, let circles be described touching that described on the base in F and G; each of these circles will touch the semicircle described on the other side. Join ODF, OEG, DE. Since AB, BC are bisected in D and E, B K DE is parallel to AC, and equal to half AC, i. e. to AO or OC. In the same manner OD is parallel to BC, and OE to AB; .. ODBE is a parallelogram, and EB, i. e. EH=OD; but OF=DE, .. DH= DF, and H is a point in the circumference of the circle FHK; and being in the circumference of BHC, it will be the point of contact, since DE joins the centres. In the same manner it may be shewn that the circle GI touches the circle ABI in I. (21.) If from any point in the circumference of a circle perpendiculars be drawn to the sides of the inscribed triangle; the three points of intersection will be in the same straight line. From D any point in the circumference of the circle ABC, let DE, DF, DG be drawn perpendicular to the sides of the inscribed triangle ACB; join EF, FG; EFG is a straight line. Join AD, BD, CD. Since the angles DFB, DGB are right angles, a circle may be described about the quadrilateral figure DGBF (vi. 13.); and the angle DFG is equal to DBG. Also since the angles DFA, DEA are right angles, a circle may be described about the quadrilateral figure DFEA; whence the angles DFE, DAE are together equal to two right angles. But ADBC being a quadrilateral figure inscribed in a circle, the angle DAC is equal to DBG, i. e. to DFG; .. DFE, DFG are equal to two right angles; and EFG is a straight line. (22.) The base of a right-angled triangle not being greater than the perpendicular; if on any line drawn from the vertex to the base a semicircle be described, and a chord equal to the perpendicular placed in it, and bisected; the point of bisection will always fall within the triangle. B P H G Let ABC be a right-angled triangle, of which the side AC is not greater than BC. From B let any line BD be drawn to the base; on which describe a semicircle BCD, and in it place EF=BC, which bisect in G; the point G is within the triangle ABC. E Take O the centre of the semicircle; draw OH perpendicular to BC; join OG. Since BC is equal to EF, OH is equal to OG; and the angles at G and H being right angles, a circle described with the centre O, and radius OG, will touch BC in H, and .. G is within the angle DBC. Also since AC is not greater than BC, DC is less than BC or EF, .. EF is nearer to the centre O, than DC is; or G falls above DC and within the angle DCB. (23.) The straight line bisecting any angle of a triangle inscribed in a given circle, cuts the circumference in a point, which is equidistant from the extremities of the side opposite to the bisected angle, and from the centre of a circle inscribed in the triangle. Let ABC be a triangle inscribed in the circle ACD. Bisect the angles BAC, ABC by the lines AD, BO, which meet in 0; Ois the centre of the circle inscribed in the triangle. Join BD, DC. The lines DB, DC, DO are equal to each other. B Because the angles DAB, DAC are equal, BD=DC; and because the angle CBD-CAD=DAB, to each of these add the angle CBO or its equal ABO; and the whole angle OBD is equal to the two ABO, OAB, i. e. to BOD (Eucl. i. 32.); and .. OD=DB. (24.) The perpendicular from the vertex on the base of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle, whose diameter is the base. From C the vertex, let CO be drawn perpendicular to AB, the base of the equilateral triangle ABC; upon AB describe a circle ADB, and let DEF be an equilateral triangle inscribed in it; CO will be equal to a side of this triangle. G F B Draw DG bisecting the angle at D, and .. bisecting EF at right angles, consequently passing through the centre. Join EG. The angles ACO, ADG, being each equal to half the angle of an equilateral triangle, are equal to each other, and AOC=DEG, each being a right angle, and AC=AB=DG, .. CO=DE. (25.) If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected; the line joining the points of bisection will be trisected by the sides. Let ABC be an equilateral triangle inscribed in a circle; bisect the arcs AB, AC in D and E; join DE; it is divided into three equal parts in the points F and G. Since DE and BC cut off equal arcs BD, CE, they are parallel, and .. (Eucl. vi. 2.) AF=AG. Join BD, AE. The angle BFD=AFE, and DBF= AEF in the same segment, and BD=AE, since they subtend equal arcs; .. DF-FA. In the same manner it may be shewn that AG=GE. Now the triangle AFG being similar to ABC is equilateral, .. DF, FG, GE are all equal, and DE is trisected. (26.) If any triangle be inscribed in a circle, and from the vertex a line be drawn parallel to a tangent at either extremity of the base; this line will be a fourth proportional to the base and two sides. Let ABC be a triangle inscribed in the circle ABC; and from B let BD be drawn parallel to AE a tan gent at A; then will AC AB :: BC : BD. Produce CB to meet the tangent in E. Since the angle EAB is equal to the angle in the alternate segment ACB, and the D angle AEB is equal to CBD, .. the triangle ABE is similar to CBD, and AE AB :: CB: CD; but from similar triangles BDC, EAC, AC AE DC: DB, .. ex æquo AC AB :: CB: DB. (27.) If a triangle be inscribed in a circle, and from its vertex lines be drawn parallel to tangents at the extremities of its base, they will cut off similar triangles. Let ABC be a triangle inscribed in a circle, and AD, CE tangents at the points A and C. From B draw BF, BG respectively parallel to them; these lines will cut off the triangles ABF, CBG, which are similar. A D F G For (Eucl. iii. 32.) the angle ACB is equal to DAB, i. e. to the alternate angle ABF; and the angle BAC is equal to BCE, i. e. to CBG; whence the triangles ABF, CBG having two angles in each equal, will be equiangular and similar. COR. 1. The rectangle contained by the segments of the base adjacent to the angles will be equal to the square of either line drawn from the vertex. For if AD and CE be produced, they will meet and |