PROPOSITION XXXII. THEOREM.

In every triangle, the rectangle contained by two sides is equivalent to the rectangle contained by the diameter of the circumscribed circle, and the perpendicular let fall upon the third side.

In the triangle ABC, let AD be drawn perpendicular to BC; and let EC be the diameter of the circumscribed circle; then will

AB.AC

AD.CE.

I

For, drawing AE, the triangles ABD, AEC, are right angled, the one at D, the other at A: also the angle B-E; these triangles are therefore similar, and they give the proportion AB CE: AD: AC; and hence AB.AC=CE.AD.

B

Cor. If these equal quantities be multiplied by the same quantity BC, there will result AB.AC.BC=CE.AD.BC; now AD.BC is double of the area of the triangle (Prop. VI.); therefore the product of three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle.

The product of three lines is sometimes called a solid, for a reason that shall be seen afterwards. Its value is easily conceived, by imagining that the lines are reduced into numbers, and multiplying these numbers together.

Scholium. It may also be demonstrated, that the area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle.

B

D

For, the triangles AOB, BOC, AOC, which have a common vertex at O, have for their common altitude the radius of the inscribed circle; hence the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half the radius

F

OD; hence the area of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle.

PROPOSITION XXXIII. THEOREM.

In every quadrilateral inscribed in a circle, the rectangle of the two diagonals is equivalent to the sum of the rectangles of the opposite sides.

In the quadrilateral ABCD, we shall have

AC.BD=AB.CD+AD.BC.

we shall have

Take the arc CO=AD, and draw BO meeting the diagonal AC in I.

The angle ABD CBI, since the one has for its measure half of the arc AD, and the other, half of CO, equal to AD; the angle ADB BCI, because they are A both inscribed in the same segment AOB; hence the triangle ABD is similar to the triangle IBC, and we have the proportion AD CI: BD: BC; hence AD.BC=CI.BD. Again, the triangle ABI is similar to the triangle BDC; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc AO=DC; hence the angle ABI is equal to DBC; also the angle BAI to BDC, because they are inscribed in the same segment; hence the triangles ABI, DBC, are similar, and the homologous sides give the proportion AB : BD AI: CD; hence AB.CD=AI.BD.

Adding the two results obtained, and observing that

AI.BD+CI.BD=(AI+CI).BD=AC.BD,

AD.BC+AB.CD=AC.BD.

I

PROBLEMS RELATING TO THE FOURTH BOOK.

PROBLEM I.

To divide a given straight line into any number of equal part, or into parts proportional to given lines.

First. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefinite straight line AG; and taking AC of any magnitude, apply it five times upon AG; join the last point of division G, and the extremity B, by the straight line GB; then draw CI parallel to GB: AI will be the fifth part of the line AB; and thus, by applying AI five times upon AB, the line AB will be divided into five equal parts.

E

F

A.

D

G

For, since CI is parallel to GB, the sides AG, AB, are cut proportionally in C and I (Prop. XV.). But AC is the fifth part of AG, hence AI is the fifth part of AB,

D

M

B

Pr

Secondly. Let it be proposed to divide the line AB into parts proportional to the given lines P, Q, R. Through A, draw the indefinite line AG; make AC= QP, CD=Q, DE=R; join R the extremities E and B; and through the points C,

ENG

D, draw CI, DF, parallel to EB; the line AB will be divided into parts AI, IF, FB, proportional to the given lines P, Q, R.

IF B

For, by reason of the paral.cls CI, DF, EB, the parts AI, IF, FB, are proportional to the parts AC, CD, DE; and by construction, these are equal to the given lines P, Q, R.

PROBLEM II.

To find a fourth proportional to three given lines, A, B, C.

Draw the two indefinite lines DE, DF, forming any angle with each other. Upon DE take DA A, and DB=B; upon DF take DC=C; draw AC; and through the point B, draw BX parallel to AC; DX will be the fourth proportional required; for, since BX is parallel to AC, we have the proportion DA: DB:: DC: DX; now the first three terms of this proportion are equal to the three given lines: consequently DX is the fourth proportional required.

F

A

E

B

D

PROBLEM III.

I

PROBLEM IV.

A

B

CI

Cor. A third proportional to two given lines A, B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines A, B, B.

A

C

To find a mean proportional between two given lines A and B.

Upon the indefinite line DF, take DE=A, and EF-B; upon the whole line DF, as a diameter, describe the semicircle DGF at the point E, erect upon the diameter the perpendicular EG meeting the circumference in G; EG will be the mean proportional required.

For, the perpendicular EG, let fall from a point in the circumference upon the diameter, is a mean proportional between DE EF, the two segments of the diameter (Prop. XXIII. Co.); and these segments are equal to the given lines A and B.

BH
AH

E

To divide a given line into two parts, such that the greater part shall be a mean proportional between the whole line and the other part.

Let AB be the given line. At the extremity B of the line AB, erect the perpendicular BC equal to the half of AB; from the point C, as a centre, with the radius CB, describe a semicircle; draw AC cutting the circumference in D; and take AF-AD: the line AB will be divided at the point F in the manner required; that is, we shall have AB : AF :: AF : FB.

A

F

B

For, AB being perpendicular to the radius at its extremity, is a tangent; and if AC be produced till it again meets the circumference in E, we shall have AE AB : : AB : AD (Prop. XXX.); hence, by division, AE-AB: AB : : AB— AD: AD. But since the radius is the half of AB, the diameter DE is equal to AB, and consequently AE-AB-=AD=AF; also, because AF AD, we have AB-AD=FB; hence AF: AB :: FB: AD or AF; whence, by exchanging the extremes for the means, AB AF :: AF : FB.

Scholium. This sort of division of the line AB is called di vision in extreme and mean ratio: the use of it will be perceived in a future part of the work. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB-DE, we have AE: DE :: DE: AD.

PROBLEM V.

Through a given point, in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal.

с

Let BCD be the given angle, and A the given point. Through the point A, draw AE parallel to CD, make BE-CE, and through the points B and A draw BAD; this will be the line required.

For, AE being para to CD, we have BE EC BA: AD, but E=FC; therefore BA=AD.

B/

E