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PROPOSITION B.-THEOREM.

If an angle of a triangle be bisected by a straight line which likewise 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 straight line bisecting the angle.

LET a b c be a triangle, and let the angle ba c be bisected by the straight line ad; the rectangle ba, a c is equal to the rectangle b d, dc, together with the square of a d

:

b

d

Describe the circle (iv. 5) a cb about the triangle, and produce a d to the circumference in e, and join ec. Then because the angle bad is equal to the angle cae, and the angle abd to the angle (iii. 21) a e c, for they are in the same segment; the triangles a bd, a e c, are equiangular to one another therefore as ba to ad, so is (vi. 4) ea to a c, and consequently the rectangle ba, a c is equal (vi. 16) to the rectangle ea, a d that is (ii. 3), to the rectangle ed, da, together with the square of a d: but the rectangle ed, da is equal to the rectangle (iii. 35) bd, dc. Therefore the rectangle ba, a c is equal to the rectangle bd, dc, together with the square of a d. Wherefore, if an angle, &c. Q. E. D.

e

PROPOSITION C.-THEOREM.

If from any angle of a triangle a straight line be drawn perpendicular 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 from the angle a to the base bc; the rectangle ba, a c is equal to the rectangle contained by ad, and the diameter of the circle described

about the triangle.

Describe (iv. 5) the circle acb about the triangle, and draw its diameter a e, and join ec: because the right angle bda is equal b (iii. 31) to the angle e ca in a semicircle, and the angle abd to the angle aec in the same segment (iii. 21); the triangles ab d, a ec are equiangular therefore as (vi. 4) ba to a d, so is ea to a c; and consequently the rectangle ba, ac is equal (vi. 16) to the rectangle e a, a d. If therefore from an angle, &c. Q. E. D.

d

L

PROPOSITION D.-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 join a c, bd ; the rectangle contained by a c, bd is equal to the two rectangles contained by a b, cd, and by a d, bc.*

Make the angle a be equal to the angle dbc: add to each of these the common angle ebd, then the angle a b d is equal to the angle ebc: and the angle bda is equal (iii. 21) to the angle bce, because they are

b

in the same segment: therefore the triangle a bd is equiangular to the triangle bce: wherefore (vi. 4), as bc is to ce, so is bd to da; and consequently the rectangle bc, ad is equal (vi. 16) to the rectangle bd, ce again, because the angle a be is equal to the angle dbc, and the angle (iii. 21) bae to the angle bd c, the triangle a be is equiangular to the triangle bcd: as therefore ba to ae, so is bd to dc; wherefore the rectangle ba, dc is equal to the rectangle bd, ae but the rectangle bc, ad has been shewn equal to the rectangle bd, ce; therefore the whole rectangle ac, bd (ii. 1) is equal to the rectangle ab, dc, together with the rectangle ad, bc. Therefore, the rectangle, &c. Q. E. D.

d

EXERCISES ON BOOK VI.

SECT. I.-PROBLEMS.

1. GIVEN the base, the vertical angle, and the ratio between the two sides, to construct the triangle.

2. Given the vertical angle, the perpendicular from it to the base, and the ratio between the segments and the base on each side of the perpendicular, to construct the triangle.

3. Given a scalene triangle, to construct an isosceles triangle equal to it, and with an equal vertical angle.

4. Given a finite straight line, to describe on it a right-angled triangle whose sides shall be continual proportionals.

5. Given an angle and a point, to draw from the point a straight line cutting the lines that contain the angle, in such a way that the parts be

*This is a Lemma of Cl. Ptolemæus, in page 9 of his μeyaλN OVVTAĘIS.

tween the vertex and the points of intersection shall have to each other a given ratio.

6. Given a straight line, to divide it into two parts such that the square of one of the parts shall have a given ratio to the rectangle contained by the whole line and the other part.

7. Given the first and fourth of four proportional lines and the difference between the second and third, to find the second and third.

8. Given a rectilineal figure, to describe a square which shall have a given ratio to it.

9. Given a trapezium in which the sides about one angle are equal, and the sides about the opposite angle also equal, to inscribe in it a

square.

10. Given two similar rectilineal figures, to find a third similar rectilineal figure which shall be a mean proportional between them.

11. Given a segment of a circle, to divide it into two parts so that the chords of these parts shall have a given ratio to each other.

12. To trisect a given circle.

SECT. II. THEOREMS.

13. If a straight line be drawn from the vertex of an isosceles triangle at right angles to one of the equal sides and produced until it meets the base produced, either of the equal sides is a mean proportional between the base and one-half of the base produced.

14. If the three sides of a triangle be bisected and straight lines drawn from the points of bisection to the opposite angles, they shall intersect each other in the same point.

15. If two or more lines meet three parallel lines, they are cut proportionally.

16. If a straight line touch two circles that also touch each other, the mean proportional between the diameters of the circles is that part of the line lying between the points of contact.

17. If parallelograms be equiangular, the ratio between them is the same as the ratio between the rectangles contained by the sides about equal angles in each.

THE END.

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