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For let the triangle a b c be divided into two triangles a cd, dcb, by making the angle a cd=dc b (by postulate 4.) then because a c=b c, and c d common (by the last) the triangle a d c=dcb; and therefore the angle a=b. Q. E. D.

Cor. Hence if from any point in a perpendicular which besects a given line, there be drawn right lines to the extremities of the given one, they with it will form an isosceles triangle,




The angle BCD at the centre of a circle ABED is double the angle BAD at the circumference, stand, ing upon the same are BED. fig. 25,

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Through the point A, and the centre C, draw the line ACE: then the angle ECD=CAD+CDA ; (by theo. 4.) but since AC=CD being radii of the same circle, it is plain (by the preceding lemma) that the angles subtended by them will be also equal, and that their sum is double to either of them, that is, DAC+ ADC is double to CAD, and therefore ECD is double to CAD; after the same manner BCE is double to CAB, wherefore, BCE+ECD, or BCD is double to BAC+CAD or to BAD. Q. E. D.

Cor. Hence an angle at the circumference is measured by half the arc it subtends or stands on.

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Cor. 3. Hence an angle in a segment greater than a semicircle is less than a right angle; thus ADB is measured by half the arc AB, but as the arc AB is less than a semicircle, therefore half the arc AB, or the angle ADB is less than half a semicircle, and consequently less than a right angle. fig. 26.

fig. 27


Cor. 4. An angle in a segment less than a semicircle, is greater than a right angle, for since the arc AEC is greater than a semicircle, its half which is the measure of the angle ABC, must be greater than half a semicircle, that is, greater


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Let the lines CA and CB be drawn from the center to the extremities of the chord, then since CA= CB, the angle CAB=CBA (by the lemma.) But the triangles ADC, BDC are right angled ones, since the line CD is a perpendicular; and so the angle ACD =DCB; (by cor. 2. theo. 5.) then have we AC, CD, and the angle ACD in one triangle ; severally equal to CB, CD, and the angle BCD in the other; therefore (by theo. 6.) A=DB. Q. E. D.



Cor. Hence it follows, that any line bisecting a chord at right angles, is a diameter; for a line drawn from the centre perpendicular to a chord, bisects that chord at right angles; therefore, conversely, a line bisecting a chord at right angles must pass through the centre, and consequently be a diameter.


If from the centre of a circle ABE there be drawn perpendicular CD on the chord AB, and produced till it meets the circle in F, that line CF, will bisect the arc AB in the point F. fig. 29.

Let the lines AF and BF be drawn, then in the triangles ADF, BDF: AD=BD (by the last ;) DF is common, and the angle ADF=BDF being both right, for CD or DF is a perpendicular. Therefore (by theo. 6.) AF-FB; but in the same circle, equal lines are chords of equal arcs, since they measure them (by def. 19.) whence the arc AF=FB, and so AFB is bisected in F, by the line CF.

Cor. Hence the sine of an arc is half the chord of twice that arc. For AD is the sine of the arc AF and AF is half the arc, and AD half the chord AB (by theo. 8.) therefore the cor. is plain.

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Let the circle ADB be drawn thro' the points A, B, BKD (by cor. 1. theo. 7.) viz. the chord of BK is the measure of the angle BAD; therefore (by cor. to the last) BE the half of BD is the sine of BAD; the same way may be proved, that half of AD is the sine of ABD, and the half of AB the sine of ADB. Q. E, D.

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I If a right line GH cut two other A right lines AB, CD, so as to make

C the alternate angles AEF, EFD

H equal to each other, then the lines AB and CD will be parallel. fig. 22.



If it be denied that AB is parallel to CD, let IK be parallel to it; then IEF=(FED)=AEF (by part 2. theo. 3.) a greater to a less, which is absurd, whence IK is not parallel ; and the like we can prove

of all other lines but AB; therefore AB is parallel to CD. Q. E. D.


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If two equal and parallel lines AB, CD be joined by two other lines AD, BC, those shall be also equal and parallel. fig. 3.


Let the diameter or diagonal BD, be drawn and we will have the two triangles ABD, CBC, whereof AB in one is = to CD in the other, DB common to both, and the angle ABD=CDB (by part 2. theo. 3.) therefore (by theo. 6.) AD=CB, and the angle CBD=

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