Elements of geometry and trigonometry . the radius CH perpendicu-lar to the chord MP. It will, at thesame time be perpendicular to NQ(Book ); there fore, thepoint H will be at once the middle ofthe arc M HP, and of the arc NHQ(Prop. VI.) ; therefore, we shall havethe arc MH = HP, and the arc NH=:HQ; and therefore MH—NH=:HP—HQ ; in other words,MN^PQ. Second. When, of the two paral-lels AB, DE, one is a secant, theother a tangent, draw the radius CHto the point of contact H ; it will beperpendicular to the tangent DE(Prop. IX.), and also to its parallelMP. But, since CH is perpend
Elements of geometry and trigonometry . the radius CH perpendicu-lar to the chord MP. It will, at thesame time be perpendicular to NQ(Book ); there fore, thepoint H will be at once the middle ofthe arc M HP, and of the arc NHQ(Prop. VI.) ; therefore, we shall havethe arc MH = HP, and the arc NH=:HQ; and therefore MH—NH=:HP—HQ ; in other words,MN^PQ. Second. When, of the two paral-lels AB, DE, one is a secant, theother a tangent, draw the radius CHto the point of contact H ; it will beperpendicular to the tangent DE(Prop. IX.), and also to its parallelMP. But, since CH is perpendicularto the chord MP, the point H must bethe middle of the arc MHP () ; therefore the arcs MH, HP, in-cluded between the parallels AB, DE, are equal. Third, If the two parallels DE, IL, are tangents, the oneat H, the other at K, draw the parallel secant AB ; and, fromwhat has just been shown, we shall have MHr^zHP, MK=KP;and hence the whole arc HMK==HPK. It is farther evidenttbat each of these arcs is a BOOK III. PROPOSITION XI. THEOREM. If two circles cut each other in two pointsj the line which passesthroui^h their centres, will be perpendicular to the chord whichjoins the points of intersection, and will divide it into t^oequal parts. For, let the line AB join the points of intersection. It willbe a common chord to the two circles. Now if a perpendicular
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