Elements of geometry and trigonometry . hemselves equal ; and, consequently, the angle ACD is equalEOG (Book 1. Prop. X.). Now, placing the semicircle ADBon its equal EGF, since the angles ACD, EOG, are equal, it isplain that the racius CD will fall on the radius OG, and thepoint D on the point G ; therefore the arc AMD is equal to thearc ENG. PROPOSITION V. THEOREM. \In the same circle, or in equal circles, a greater arc is subtendedby a greater chord, and conversely, the greater chord subtendsthe greater arc. Let ihe arc AH be greater thanthe arc AD ; then will the chord AHbe greater than th
Elements of geometry and trigonometry . hemselves equal ; and, consequently, the angle ACD is equalEOG (Book 1. Prop. X.). Now, placing the semicircle ADBon its equal EGF, since the angles ACD, EOG, are equal, it isplain that the racius CD will fall on the radius OG, and thepoint D on the point G ; therefore the arc AMD is equal to thearc ENG. PROPOSITION V. THEOREM. \In the same circle, or in equal circles, a greater arc is subtendedby a greater chord, and conversely, the greater chord subtendsthe greater arc. Let ihe arc AH be greater thanthe arc AD ; then will the chord AHbe greater than the chord AD. For, draw the radii CD, CH. Thetwo sides AC, CH, of the triangleACH are equal to the two AC, CD,of the triangle ACD, and the angleACH is greater than ACD ; hence, the third side AH is greater than the third side AD (Book I. Prop. IX.) ; there- IC fore the chord, which subtends the greater arc, is the , if the chord AH is greater than AD, it will follow,on comparing the same triangles, that the angle ACH is. BOOK III. 45 greater than ACD (Bk. I. Prop. IX. Sch.) ; and hence thatthe arc AH is greater than AD ; since the whole is greaterthan its part. Scholium. The arcs here treated of are each less than thesemicircumference. If they were greater, the reverse pro-perty would have place ; for, as the arcs increase, the chordswould diminish, and conversely. Thus, the arc AKBl) irfgreater than AKBII, and the chord AD, of the first, is lessthan the chord AH of the second. PROPOSITION VI. THEOREM. Tke radius which is perpendicular to a chord, bisects the chord,and bisects also the subtended arc of the chord. Let AB be a chord, and CG the ra-dius perpendicular to it : then will AD =DB, and the arc AG-=GB. For, draw the radii CA, CB. Thenthe two right angled triangles ADC,CDB, will have AC = CB, and CD com-mon ; hence, AD is equal to DB (BookI. Prop. XVII.). Again, since AD, DB, are equal, CGis a perpendicular erected from the mid-dle of AB ; hence every poin
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