We have, also, under the same article, found the area to.be A = { (a + b + c ) ( − a + b + c) (a = b+c) (a + b − c ) } *.[4] We know [B. IV, Prop. xxvII] that the diameter of the circumscribing circle, multiplied into either perpendicular, is equal to the product of the sides containing the angle from which the perpendicular is drawn. Hence Substituting for P, its value already found, we have We will now seek the radius r, of the escribed circle, whose centre is at D. Area DBA + DCA - DBC is evidently equal to the area ABC. Area DBA is equal to the base AB multiplied by half the perpendicular drawn from D upon AB produced hence area DBA = c × 1 r, = cr1. In a similar way, we find area DCA = br1, and area DBC = ar1. Since any side of a triangle, multiplied by the perpendicular which meets it from the opposite angle, gives double the area of the triangle, we have 2▲ = a a P1, or Taking the product of [13], [14] and [15], we have 843 Taking the product of [17], [18] and [19], we have [17] [18] [19] [20] Extracting the cube root of the product of [16] & [20], Dividing [20] by the square of [16], we find [21] R3 P3 P3 P3 1, or RP,P,P1 = 2 A2. [22] By taking the continued product of [7], [9], [10] and [11], we have (a+b+c)(−a+b+c)(a−b+c)(a+b−c) 16 [23] By multiplying [5] and [7] together, or [7] and [21], By combining the values of [9], [10] and [11], by two and two, we find r1r2+r1r3+r2r3 = (a+b+c)". [26] Any of the foregoing expressions, when properly translated into common language, leads to a theorem. We will translate some of the most interesting ones. Equation [12] gives the following THEOREM. The reciprocal of the radius of the inscribed circle is equal to the sum of the reciprocals of the radii of the three escribed circles. Ꮓ |