ABC and CDF will be constructed in a similar manner, if CD

:

be not less than a third of BC.

G

If DC, one of the segments, be less than a third of BC, the triangle BDE is constructed as before, but the rest of the preceding solution fails, as the second parallel would fall above the triangle. In this case, take EG equal to EB; then, if DG be joined, the triangle ABC is trisected by DE, DG.

B

E

D C

Schol. It is easy to see that this method may be readily extended to the division of a triangle into more equal parts than three, or into parts proportional to given magnitudes, by straight lines drawn from a given point in one of the sides.

PROP. XXII. PROB.

To divide a given quadrilateral into two parts in a given ratio, by a straight line parallel to one of its sides.

Let it be required to divide the quadrilateral ABCD into two parts in a given ratio, by a straight line parallel to the side BC.

D

L

M

C

G K

B

F H

Produce BA, CD to meet in E, and (I. 45.) describe a rectangle equal to ABCD. Let FG, a side of that rectangle, be divided in H, so that FH is to HG as the required part next BC to the other part; and to one of the adjoining sides of the same rectangle, apply (I. 44.) a rectangle equal to the triangle ADE. Let GK be a side of the latter rectangle; and (APP. IV. 18.) draw LM parallel to BC, dividing the triangle EBC, so that ELM: LBCM :: KH : HF; LM obviously divides the quadrilateral in the given ratio.

If BA, CD be parallel, the foregoing solution fails. In that case, the solution is effected simply by describing (I. 44.) on BC a parallelogram equal to the one on FH, and having two of its sides coinciding with BA, CD.

PROP. XXIII. THEOR.

Ir the sides of a right-angled triangle be continual proportionals, the hypotenuse is divided in extreme and mean ratio by the perpendicular to it from the right angle; and the greater segment is equal to the less or remote side of the triangle.

Let ABC be a triangle right-angled at A, and let AD be per

pendicular to BC; then if CB: BA :: BA: AC, BC is divided in extreme and mean ratio in D, and BD is equal to AC.

B

A

D

For (hyp.) CB : BA :: BA : AC, and (VI. 8. cor.) CB: BA :: BA: BD; therefore BA: AC:: BA: BD, and AC is equal to BD. Again, (VI. 8. cor.) BC: CA :: CA: CD, or BC: BD:: BD: DC, and therefore (VI. def. 3.) BC is divided in extreme and mean ratio in D.

Schol. Conversely, if BC: BD :: BD : CD, and if BAC be a right angle, and DA perpendicular to BC; CB: BA :: BA: AC, and BD is equal to AC. For (hyp.) CB: BD :: BD: CD, and (VI. 8. cor.) CB : CA :: CA: CD; wherefore BD is equal to CA3, each (VI. 17.) being equal to the rectangle BC.CD, and therefore BD is equal to CA. Again, (VI. 8.) ČB : BA :: BA:

BD or AC.

PROP. XXIV. PROB.

GIVEN the angles and diagonals of a parallelogram; to construct it.

On one of the diagonals describe a segment of a circle containing an angle equal to the given angle at either extremity of the other; from the middle point of this diagonal as centre, with half the other diagonal as radius, describe an arc cutting the arc of the segment; through the extremities of the first diagonal draw four straight lines, two to the intersection of the arcs, and two parallel to these: the parallelogram thus formed is easily proved to be the one required.

PROP. XXV. PROB.

GIVEN the vertical angle of a triangle, and the radii of the circles inscribed in the parts into which the triangle is divided by the perpendicular; to construct the triangle.

Take any straight line ABC, and through any point B draw the perpendicular BD; make BA, BC equal to the given radii, and let E, F be the angular points, remote from B, of squares described on AB, BC; join EF, and on it describe the segment EDF, containing an angle equal to half the given vertical angle; let the perpendicular cut the arc EDF in D, and join DE, DF; draw DG, DH making the angles EDG, FDH respectively equal to EDB, FDB: DGH is the required triangle.

E

F

G

A

B C

H

D

E

F

B C H

For (I. 26.) perpendiculars drawn from E to DB, DG are equal, and each of them is equal (const.) to the perpendicular from E to GB. Each of them therefore is equal to the given radius AB, and a circle described from E at the distance of one of these is inscribed in the triangle DGB. In the same manner it would be shown that a circle described from F as centre, with the other given radius, would be inscribed in DBH. Hence, since the angle GDH is double of EDF, GDH is equal to the given vertical angle, and the triangle GDH answers the conditions of the question. Schol. The preceding solution is strictly in accordance with the enunciation, taken in its limited sense. There will be interesting variations, however, if we regard the given circles, not merely as inscribed, but as those which touch all the sides of each of the right-angled triangles, either internally or externally. These variations will be obtained, by giving the squares on the radii every possible variety of position in the four right angles formed by the intersection of AC, DB; and the solution will obtain complete generality by taking into consideration both the points in which BD cuts the circle of which EF is a chord.

PROP. XXVI. PROB.

To bisect a given rectangle by two straight lines parallel to two of its adjacent sides, and equally distant from them.

Let ABCD be the rectangle, and AC one of its diagonals: find (IV. 4.) E the centre of the circle inscribed in the triangle ABC, and through E draw EKF, ELG parallel to BA, BC: these bisect the rectangle.

A F

B

K H

D

G

E

L

Draw EH perpendicular to AC. Then the triangles AFK, EHK are equiangular, and the side AF, being equal to the radius of the inscribed circle, is equal to EH. Hence (I. 26.) the triangles AFK, EHK are equal and it would be proved in a similar manner, that CGL, LHE are equal; and consequently the two AFK, CGL are equal to KEL. From the triangle ABC take KEL, and to the remainder add AFK, CGL; and from ADC take AFK, CGL, and to the remainder add KEL: then (I. ax. 3. and 2.) the figure ABCGEF is equal to the rectangle FG; that is, the given rectangle is bisected by EF, EG.

PROP. XXVII. PROB.

LET the four points A, B, C, D, lying in the same straight line, be given in position; to find another

point, such that if straight lines be drawn from it to A, B, C, D, the angles at that point subtended by AB, and CD, may be equal to one another, and the one subtended by BC may be equal to a given angle.

Find mean proportionals between AB, BD, and AC, CD; and on the base BC, describe (APP. IV. 6. cor.) the triangle BEC, having the angle BEC equal to the given angle, and BE to EC as the first mean to the second: E is the re

quired point.

E

D

A

B

F

Join EA, ED, and about the triangle AED describe a circle, meeting EB, EC produced in F and G; and join FG. Then (const. and VI. 17. and 22.) BE: CE2 :: AB.BD: AC.CD, or (III. 35. and alternately) BE : FB.BE:: CE: GC.CE. Hence (VI. 1.) BE: BF:: CE : CG, and (VI. 2.) BC, FG are parallel; wherefore (III. 26. cor.) the arcs AF, GD, and (III. 27.) the angles AEF, DEG are equal.

Schol. The method of solution will be almost exactly the same, if the points A and D lie between B and C.

PROP. XXVIII. THEOR.

THE area of a triangle ABC is equal to half the continued product of two of its sides, AB, BC, and the sine of their contained angle B, to the radius 1.

Daw the perpendicular AD. Then (APP. III. 1. cor.) AD= AB X sin B. Multiply by BC, and take half the

product; and (I. 46. cor. 6.) we have the area equal to ABX BC X sin B.

Cor. Hence (I. 34.) the area of a parallelogram is equal to the continual product of two contiguous sides, and the sine of the contained angle.

D C

• Analysis. Suppose E to be the required point, and join EA, ED; describe the circle AED, and draw EBF, ECG, and FG. Then, because the angles AEF, DEG are equal, the arcs AF, DG are (III. 26.) equal, and (III. 26. note to cor.) FG is parallel to AD. Hence (VI. 2.) EB: BF:: EC: CG, and therefore (VI. 1. and III. 35.) EB2: EB. BF or AB.BD:: EC2 EC.CG or AC. CD, or, alternately, EB2: EC2 :: AB.BD: AC.CD: whence (VI. 22.) EB is to EC as a mean proportional between AB, BD to a mean between AC, CD; and since AB, BD, AC, CD are given, these means are given. In the triangle EBC, therefore, the base, the ratio of the sides, and the vertical angle are given; whence, by the corollary to the sixth proposition of this book, the triangle, and consequently the point E are given.

:

PROP. XXIX. THEOR.

LET a, b, c be the sides of a triangle, and s half their sum; the area is equal to the square root of the continual product of s, s—a, sb, and s—c.

It was proved in the preceding book, in the second corollary to the ninth proposition, that the sine of twice any angle is twice the product of the sine and cosine of the angle. Hence, by multiplying together the values of sinA and costA, given in the corollaries to the sixth and seventh propositions of the same book, and 2√s (s—a) (s—b) (s—c)

bc

doubling the result, we get sinA = Now, by the preceding proposition, the area of a triangle is found by multiplying the sine of one of its angles by the sides containing it, and taking half of the product: multiplying, therefore, the value now found for sin A, by be, and taking half the product, we find the area to be √s (s — a) (s — b) (s — c).

PROP. XXX. PROB.

GIVEN three points A, B, C, and the angles which straight lines drawn from them to a fourth point, make with one another; to find the position of that point. Join AB, BC, AC; and on BC describe the segment BDC, containing an angle equal to the one formed. by straight lines drawn from B and C to the required point: then, if a segment were described on AB containing an angle equal to the one formed by lines drawn from the required point to A and B, the arcs of the segments would intersect one another in D, which is the required point. For if DA, DB, DC be joined, the angles BDC, BDA are obviously of the given magnitudes, as they are angles in segments described so as to contain those angles.

C

Method of Computation. Complete the circle BDCE, and let AD cut it in E; join BE, CE. Now, since the positions of the points A, B, C are given, the sides of the triangle ABC, and consequently its angles are given. Then, in the triangle BCE, the side BC is given, as are also the angles EBC, ECB, which (III. 21.) are respectively equal to EDC, EDB. Hence BE may be computed; and then, in the triangle ABE, the sides