Cor.-A circle, described about the hypothenuse [BC], of a right angled triangle [ABC], passes through the right angle [BAC]; for, if it cut the right line [BF] in any other point but [A], the angle [BAC] would be greater or less, than the angle formed at the intersection, by right lines drawn from it to the extremes of the hypothenuse BC [16. 1], which angle so formed being, by this prop. a right one; the angle [BAC] would be greater or less than a right angle, contrary to the hypothesis.

PROP. XXXII. THEOR.

If a right line (EF) touch a circle (ABCD), and from the contact (B), a right line (BD) be drawn cutting the circle; the angles made by the tangent and cutting line, are equal to the angles in the alternate segments.

If the cutting line pass through the centre, the angles are equal, being right angles [18 and 31. 3].

E

B

If not, from the contact B, draw BA at right angles to EF [11. 1], and, having taken any point C, in the circumference BCD, join AD, DC, CB; and because BA is perpendicular to the tangent EF, the centre of the circle is in BA [19. 3], and the angle BDA in a semicircle is a right angle [31. 3], and, therefore, in the triangle ADB, the other two angles BAD, ABD, are equal to a right angle [32. 1], and therefore to the right angle ABF; taking from each the common angle ABD, the angle DBF is equal to the angle BAD in the alternate segment. And, because, in the quadrilateral figure ABCD inscribed in a circle, the opposite angles BAD, BCD are equal to two right angles [22. 3], and therefore to the angles DBF, DBE, which are also equal to two right angles [13. 1]; taking away the equal angles BAD, DBF, the remaining angle DBE is equal to the remaining angle DCB in the alternate segment.

PROP. XXXIII. PROB.

On a given right line (AB), to describe a segment of a circle, which may receive an angle, equal to a given rectilineal angle (C).

First.-Let the given angle C be a right one, [see fig. 1]. Bisect the given right line AB in F [10. 1], from the centre F, at the distance FA, describe the semicircle AHB; the angle AHB in a semicircle is equal to the right angle C [31. 3].

Secondly.-Let the angle C not be a right one (see fig. 2 and 3), and at the point A, with the right line AB, make the angle BAD equal to C [23. 1], and from the point A, draw AE at right angles to AD (11. 1); bisect AB in F, from F draw FG perpendicular to AB (11. 1), and join GB; and because, in the triangles AFG, BFG, AF is equal to FB, FG common, and the angles

AFG, BFG equal [Theor. at 11. 11, the bases AG, BG are equal [4. 1], and the circle described from the centre G, at the distance GA, passes through B; let this circle be AHB; and, because AD is drawn from the extremity of the diameter AE, perpendicular to it, AD touches the circle [Cor. 1. 16. 3], and therefore the angle BAD, or which is equal [Constr.], the angle C, is equal to the angle in the alternate segment AHB [32. 3]; and so a segment AHB is described on the given right line AB, which may receive an angle, equal to the given angle C, as was required to be done,

PROP. XXXIV. PROB.

From a given circle receive an angle,

ABC), to cut off a segment, which may equal to a given rectilineal angle (D),

Draw EF touching the circle in any point B (17.3), and at the point B with the right line BF, make the angle FBC equal to D [23. 1]: and, because EF is a tangent to the circle, the angle FBC, or, which is equal (Constr.), the angle D, is equal to the angle in the alternate segment BAC (32. 3); and there

fore, there is cut off from the given circle, a segment BAC, reCving an angle equal to the given angle D, as was required to be done,

PROP. XXXV. THEOR.

If two right lines (AC, BD), inscribed in a circle (ABCD), cut each other, the rectangle (AEC), under the segments of one, is equal to the rectangle (BEL), under the segments of the other.

Case 1.-If both pass through the centre; AE, EC, BE, ED being all equal (Def. 10. 1), the rectangle AEC is equal to the rectangle BED (Cor. 3. 34. 1).

Case 2.-If one of them BD, passing through the centre F, cut the other AC, not passing through the centre, perpendicularly, join AF; and because BD is cut. equally in F, and unequally in E, the rectangle DEB with the square of EF, is equal to the square of FB (5. 2), or of FA, and therefore, to the squares of AE and EF (47. 1); taking away

E

the common square of EF, the rectangle DEB is equal to the square of AE, or AE, EC being equal (3. 3), to the rectangle AEC.

D

G

Case 3.-If one of them DB, passing through the centre F, cut the other AC, not passing through the centre, obliquely, join AF, and draw FG perpendicular to AC; and since DB is divided equally in F, and unequally in E, the rectangle DEB and the square of FE, or, (the squares of FG, GE, being equal to the square of FE 47. 1,), the rectangle DEB, and the squares of FG, GE, are equal to the square of FB (5. 2), or FA, or, which is equal (47. 1), to the squares AG, GF; taking away the common square of FG, the rectangle DEB with the square of GE, is equal to the square of AG; but, because FG is perpendicular to AC, AC is bisected in G (3. 3), therefore the rectangle AEC with the square of GE, is equal to the square of AG [5. 2]; wherefore the rectangle DEB with the square of GE, is equal to the rectangle AEC with the square of GE; taking away the common square of GE, the rectangle DEB is equal to the rectangle AEC.

Case 4.-But if neither of the right lines AC, BD pass through the centre, find the centre F, and through E, draw the diameter HG; then the rectangles AEC, DEB, being each of them equal to the rectangle HEG (case 3 of this), are equal to each other.

H

F

Otherwise. (see all the figures to this Prop.)

Join AB, CD; and, since the triangles AEB, DEC, having their angles at E equal (15. 1), and the angles ABE, DCE, in the same segment ABCD, also equal (21. 3), are equiangular (32. 1), the rectangles under the sides about the equal angles AEB, DEC, taken alternately, are equal (Cor. 4. 5 & 6. 2), namely, the rectangle AEC to the rectangle DEB.

PROP. XXXVI. THEOR.

If from any point (E) without a circle (CAD), two right lines (ED, EA) be drawn to it, one of which (ED) cuts it, and the other (EA) touches it; the rectangle under the whole cutting line (ED) and the external segment (EC), is equal to the square of the tangent (EA).

Case 1. If ED pass through the centre F, join AF; and, because CD is bisected in F, and CE a part added to it, the rectangle DEC with the square of FC, is equal to the square of FE (6. 2), or which is equal (47. 1), to the squares of FA and AÈ; taking away the equal squares of FC and FA, the rectangle DEC, is equal to the square of AE (Ax. 3).

Case 2. If ED do not pass through the centre F, draw FG perpendicular to it, and join FA, FC and FE; and, because DC is bisected in G (3. 3), and CE added to it, the rectangle DEC with the square of GC, is equal to the square of GE (6. 2); adding to each the square of GF, the rectangle DEC, and the squares of CG, GF, or, (the squares of CG, GF being equal to the square of FC 47. 1,), the rectangle DEC, and the square of FC, are equal to the squares

A

ཁ

of EG, GF, or, which is equal (47. 1), to the square of FE, or, which is equal (47. 1), to the squares of FA, AE; taking away the equal squares of FC, FA,' the rectangle DEC is equal to the square of AE (Ax. 3).

Otherwise, (See both figures to this prop.)

Join DA, AC; and, since the triangles ECA, EAD, having the angle AED common, and the angles EAC, EDA equal (32. 3), are equiangular (32. 1), the rectangles under the sides about the common angle AED, taken alternately, are equal (Cor. 4. 5 and 6. 2), namely, the rectangle CED to the square. of EA.