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But the arc BE is less than the arc BF, therefore the angle BGE is less than the angle BGD, which is impossible.
Therefore, no other arc but BD can be a fourth proportional to the angles BGC, BGD, and the arc BC.
Also because the angles (see first figures) BGC, EHF at the centres, are double (III. 16) the angles BAC, EDF, at the circumferences, therefore the angles BAC, EDF, are also proportional to the arcs BC, EF, on which they stand. So also with the sectors; for the lines drawn from the points of division will divide the sectors into equal portions, as may
be proved by applying one to another. Therefore, by a similar demonstration the sectors may be proved proportional to the arcs on which they stand.
Scholium.--An angle at the centre is said to be measured by the arc which subtends it. The angles at the centre being together equal to four right angles, a right angle is measured by a quadrant. The unit used is, however, the ninetieth part of a right angle for angles, and the ninetieth part of a quadrant for arcs. This unit in both cases is called a degree; its subdivisions are minutes and seconds. Hence an angle at the circumference contains half as many degrees as the arc on which it stands, and is said to be measured by the half-arc.
From this we may easily prove the following theorems:
3. The opposite angles of an inscribed quadrilateral are together 180°.
4. The chord of 60° is equal to radius.
Proposition 26. Theorem.-- If in a circle two chords cut each other, the rectangle contained by the segments of one is equal to the rectangle contained by the segments of the other.
Let the two chords AB, CD of the circle ADB cut each other in E; then AE. EB= CE. ED.
Join AC, DB; and because in the triangles AEC, DEB the angles AEC, DEB are equal to each other (I. 17), also the angles CAE, EDB, because they are in the same segment of the circle, therefore the other angles at C and B are equal (I. 30, Cor. 4), and the triangles are equiangular. Hence (V. 4), AE: EC :: DE : EB;
.. (V. 14), AE. EB= CE. ED.
Proposition 27. Theorem.—If from a point without a circle, two straight lines be drawn, one of which cuts the circle and the other touches it, the rectangle contained by the whole line and the part without the circle, will be equal to the square of the line touching the circle.
From the point A draw ABC, cutting the circle CBD, and AD touching it; then CA . AB= ADP.
Join DB, DC.
Because AD touches the circle, and DB is drawn from the point of contact, the angle ADB is equal to the angle in the alternate segment DCB; also the angle A is common to the two triangles ADB, ADC; hence they are equiangular, and (V. 4),
AC: AD :: AD: AB; .. (V. 14), AC. AB = AD.
Corollary 1.-If from a point without a circle, two secants be drawn to the opposite circumference, the rectangles of the whole lines and the parts without the circle, will be equal.
Draw A EF; then AC. AB=AF. AE, for each is equal to the square of AD.
Corollary 2.-Two tangents drawn from the same point are equal.
Corollary 3.- If AB. AC= AE. AF, then the four points B, E, F, C are on• the circumference of the same circle.
For, if not, let B, E, F, and H be on the same circle. Let the student prove an impossibility.
Theorem.—If an angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle, is equal to the rectangle contained by the segments of the base, together with the square of the line bisecting the angle.
Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; then
BA.AC-BD.DC+AD?. Describe (III. 29) the circle ABC about the triangle ABC, and produce AD to the circumference in E, and join EC
Then, because the angle BAD is equal to the angle EAC, and the angle ABD to the angle AEC, they being in the same segment, the triangles ABD, AEC are equiangular.
Proposition 29. Theorem.-If from any angle of a triangle a perpendicular be drawn to the base, the rectangle contained by the sides of the triangle, is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.
Let ABC be a triangle, and AD the perpendicular; then BA. AC - the rectangle DA and a
diameter of the circumscribed circle. Describe the circle ABC about the triangle ABC; draw the diameter AE, and join EC.
Because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same segment, the triangles ABD, AEC are equiangular.
Proposition 30. Theorem.— The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.
Let ABCD be any quadrilateral inscribed in a circle, and let AC, DB be drawn; then
Again, because the angle ABE is equal to the angle DBC, and the angle BAE to the angle BDC in the same segment, the triangle ABE is equiangular to the triangle DBC.
.. (V. 4), BA : AE :: BD :DC; .. (V. 14), BA.DC=BD.AE.
But we have shown BC.DA = BD.EC;
BA.DC+BC.DA = BD. AE+BD.EC.
But (II. 1) BD. AE+BD. EC=BD.AC;