Elements of geometry and trigonometry . lied by the samequantity BC, there will result = ; is double of the area of the triangle (Prop. VI.) ; there-fore the product of three sides of a triangle is equal to its areamultiplied hy twice the diameter of the circumscribed circle. The product of three lines is sometimes called â solid, for areason that shall be seen afterwards. Its value is easily con-ceived, by imagining that the lines are reduced into numbers,and multiplying these numbers together. Scholium. It may also be demonstrated, that the area ofa triangle is equ
Elements of geometry and trigonometry . lied by the samequantity BC, there will result = ; is double of the area of the triangle (Prop. VI.) ; there-fore the product of three sides of a triangle is equal to its areamultiplied hy twice the diameter of the circumscribed circle. The product of three lines is sometimes called â solid, for areason that shall be seen afterwards. Its value is easily con-ceived, by imagining that the lines are reduced into numbers,and multiplying these numbers together. Scholium. It may also be demonstrated, that the area ofa triangle is equal to its perimeter multiplied by half the radiusof the inscribed circle. For, the triangles AOB, BOC,AOC, which have a commonvertex at O, have for their com-mon altitude the radius of theinscribed circle ; hence the sumof these triangles will be equalto the sum of the bases AB, BC,AC, multiplied by half the radius OD ; hence the*^ area of the triangle ABC is equal to theperimeter multiplied by half the radius of the inscribed BOOK IV, 9t PROPOSITION XXXIII. THEOREM. In every quadrilateral inscribed in a circle^ the rectangle of thetwo diagonals is equivalent to the stLtn of the rectangles of theopposite sides. In the quadrilateral ABCD, we shall - + Take the arc CO=AD, and draw BOmeeting the diagonal AC in I. The angle ABD = CBI, since the onehas for its measure half of the arc AD,and the other, half of CO, equal to AD ;the angle ADB = BCI, because they areboth inscribed in the same segmentAOB ; hence the triangle ABD is similarto the triangle IBC, and we have theproportion AD : CI : : BD : BC ; hence ^, the triangle ABl is similar to the triangle BDC ; for thearc AD being equal to CO, if OD be added to each of them,we shall have the arc AO=:DC ; hence the angle ABI DBC ; also the angle BAI to BDC, because they are in-scribed in the same segment ; hence the triangles ABI, DBC,are similar, and the homologous sides give
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