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524. INVERSE. Since, by 502, a sect can be cut at only one point internally in a given ratio, and at only one point externally in a given ratio, therefore, by 32, Rule of Identity, if one side of a triangle is divided internally or externally in the ratio of the other sides, the line drawn from the point of division to the opposite vertex bisects the interior or exterior angle.

525. When a sect is divided internally and externally into segments having the same ratio, it is said to be divided har monically.

526. COROLLARY. The bisectors of an interior and exterior angle at one vertex of a triangle divide the opposite side harmonically.

THEOREM XII.

527. If a sect, AB, is divided harmonically at the points P and Q, the sect PQ will be divided harmonically at the points A and B.

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HYPOTHESIS. The sect AB divided internally at P, and externally at Q, so that

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528. The points A, B, and P, Q, of which each pair divide harmonically the sect terminated by the other pair, are called four Harmonic Points.

III. Rectangles and Polygons.

THEOREM XIII.

529. If four sects are proportional, the rectangle contained by the extremes is equivalent to the rectangle contained by the

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PROOF.

Then rectangle ad = bc.

On a and on b construct rectangles with altitude = c.

On c and on a construct rectangles of altitude a.

Then

a b :: ac : bc, and c: d :: ac : ad.

(504. Rectangles of equal altitudes are to each other as their bases.)

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530. INVERSE. If two rectangles are equivalent, the sides of the one will form the extremes, and the sides of the other the means, of a proportion.

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PROOF. Since ad = bc, therefore, by 489, ac : bc :: ac : ad; but ac: bc :: a: b, and ac ad :: c: d,

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531. COROLLARY. If three sects are proportional, the rectangle of the extremes is equivalent to the square on the mean.

THEOREM XIV.

532. If two chords intersect either within or without the circle, the rectangle contained by the segments of the one is equivalent to the rectangle contained by the segments of the other.

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and

HYPOTHESIS. Let the chords AB and CD intersect in P.
CONCLUSION. Rectangle AP. PB = rectangle CP. PD.
PROOF. PAC = 4 PDB,

(376. Angles in the same segment of a circle are equal.)

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533. COROLLARY.

Let the point P be without the circle,

and suppose DCP to revolve about P until C and D coincide; then the secant DCP becomes a tangent, and the rectangle

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CP.PD becomes the square on PC. Therefore, if from a point without a circle a secant and tangent be drawn, the rectangle of the whole secant and part outside the circle is equivalent to the square of the tangent.

THEOREM XV.

534. The rectangle of two sides of a triangle is equivalent to the rectangle of two sects drawn from that vertex so as to make equal angles with the two sides, and produced, one to the base, the other to the circle circumscribing the triangle.

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HYPOTHESIS. ABE X CBD.
CONCLUSION. Rectangle AB. BC = DB.BE.
PROOF. Join AE. Then

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(376. Angles in the same segment of a circle are equal.)

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535. COROLLARY I. If BD and BE coincide, they bisect the angle B; therefore rectangle AB. BC = DB. BE = DB(BD + DE) = BD2 + BD. DE = BD2 + CD. DA (by 532). Therefore, when the bisector of an angle of a triangle

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