Elements of geometry and trigonometry . PROBLEM XIIL To find (he centre of a gieen circle or arc. Take three points, A, B, C, anywhere in the circumference, or thearc ; draw AB, BC, or suj)pose t hemto be drawn ; bisect those two lines|jy the perpendiculars DE, FG :the point O, where these perpen-diculars meet, w ill be the centresf)ught (Prop. W. Sch.). Sclioliujn. The samci constnic-tion serves f^r making a circum-fcrcnrc pass through three given points A, B, C ; and also fordescribuig a circumli rence, in which, a given triangle ABGshall be 64 GEOMETRY. PROBLEM a giv
Elements of geometry and trigonometry . PROBLEM XIIL To find (he centre of a gieen circle or arc. Take three points, A, B, C, anywhere in the circumference, or thearc ; draw AB, BC, or suj)pose t hemto be drawn ; bisect those two lines|jy the perpendiculars DE, FG :the point O, where these perpen-diculars meet, w ill be the centresf)ught (Prop. W. Sch.). Sclioliujn. The samci constnic-tion serves f^r making a circum-fcrcnrc pass through three given points A, B, C ; and also fordescribuig a circumli rence, in which, a given triangle ABGshall be 64 GEOMETRY. PROBLEM a given pointy to dram a tangent l& a given circle. If the given point A lies in the circum- 2ference, draw the radius CA, and erectAD perpendicular to it : AD will be thetangent required (Prop. IX.). If the point A lies without the circle,join A and the centre, by the straightline CA : bisect CA in O ; from O as acentre, with the radius OC, describe acircumference intersecting the given cir-cumference in B ; draw AB : this will bethe tangent required. For, drawing CB, the angle CBA be-ing inscribed in a semicircle is a rightangle (Prop. XVIII. Cor. 2.) ; thereforeAB is a perpendicular at the extremityof the radius CB ; therefore it is a tan-gent. Scholium, When the point A lies without the circle, therewill evidently be always two equal tangents AB, AD, passingthrough the point A : they are equal, because tlie right angledtriangles CBA, CDA, have the hypothenuse CA common, andthe side CB = CD; hence they are equal (Book I. Prop. XVII.);hence AD is
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