CHAPTER XIX. ON TRIANGLES AND CIRCLES. 273. To find the Area of a Triangle. The area of the triangle ABC is denoted by A. Through A draw HK parallel to BC, and through ABC draw lines AD, BK, CH perpendicular to BC. The area of the triangle ABC is half that of the rectangular parallelogram BCHK [Euc. 1. 41]. .:. ▲ = √s (s − a) (s — b) (s — c) – S................. (ii). 274. To find the Radius of the Circumscribing Circle. R a Let a circle AA'CB be described about the triangle ABC. Let R stand for its radius. Let O be the centre of the circle. Join BO, and produce it to cut the circumference in A'. Join A'C. Then the angles BAC, BA'C in the same segment are equal. And the angle BCA' in a semicircle is a right angle. Thus d, the value of each of these fractions, is the diameter of the circumscribing circle. 276. To find the radius of the Inscribed Circle. Let D, E, F be the points in which the circle inscribed in the triangle ABC touches the sides. of the circle; let r be its radius. Then ID=IE=IF=r. or, The area of the triangle ABC = area of IBC + area of ICA + area of IAB. And the area of the triangle IBC = ID. BC = r. a, .. area of ABC = {ID. BC + {IE .CA+§IF.AB ▲ = r (a+b+c) = 1r. 28 = rs. 277. A circle which touches one of the sides of a triangle and the other two sides produced is called an Escribed Circle of the triangle. 278. To find the radius of an Escribed Circle. Let an escribed circle touch the side BC and the sides AC, AB produced in the points D, E, F, respectively. Let I, be its centre, r, its radius. Then ΟΙ I‚D1=I‚E‚=I,F‚=r.. The area of the triangle ABC = area of ABIC — area of I,BC, = area of ICA + area of IAB – area of I,BC, ▲ = {I,E ̧ . CA + }IF ̧ . AB - {ID ̧. BC 1 279. Similarly if r, and r, be the radii of the other two escribed circles of the triangle ABC, then 280. The following results are often useful. With the construction of the last two articles; the lines IA, IB, IC bisect the angles A, B, C respectively; the lines IA, IB, I,C′ bisect the angles BAC, CBF, BCE, respectively; so that AII, is a straight line. A F B F (i) Two tangents drawn from an external point to touch a circle are equal; therefore AE AF; BD = BF; CE = CD; = (ii) 2AE1 = AE, + AF1 = (AC + CD1) + (AB+ BD1) = AC+ CB+BA = a+b+c= 28. (iii) 24E=AE+AF= (AC - CD) + (AB − BD) Similarly, BF = BD = (8 − b); CD=CE = (8 −c). (v) Hence, BD, CD, and CD, = BD. = |