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PROPOSITION XXX. THEOREM.

The rectangle contained by the diagonals of a quadrilatera inscribed in a circle, is equivalent to the sum of the rectangles of the opposite sides.

Let ABCD be any quadrilateral inscribed in a circle, and let the diagonals AC, BD be drawn; the rectangle AC× BD is equivalent to the sum of the two rectangles ADX BC and ABXCD.

Draw the straight line BE, making the angle ABE equal to the angle DBC. To A each of these equals add the angle EBD;

B

then will the angle ABD be equal to the angle EBC. But the angle BDA is equal to the angle BCE, because they are both in the same segment (Prop. XV., Cor. 1, B. III.); hence the triangle ABD is equiangular and similar to the triangle EBC. Therefore we have

AD: BD::CE: BC; and, consequently, AD× BC=BDxCE.

Again, because the angle ABE is equal to the angle DBC and the angle BAE to the angle BDC, being angles in the same segment, the triangle ABE is similar to the triangle DBC; and hence

consequently,

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Adding together these two results, we obtain
ADX BC+ABXCD=BDxCE+BDXAE,
which equals BD× (CE+AE), or BD × AC.
Therefore, the rectangle, &c.

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If from any angle of a triangle, a perpendicular be drawn to the opposite side or base, the rectangle contained by the sum and difference of the other two sides, is equivalent to the rectangle contained by the sum and difference of the segments of

the base.

Let ABC be any triangle, and let AD be a perpendicular drawn from the angle A on the base BC; then

(AC+AB) × (AC-AB)=(CD+DB) x (CD-DB). From A as a center, with a radius equal to AB, the short

E

G

C

B

F

er of the two sides, describe a circumference BFE. Produce AC to meet the circumference in E, and CB, if necessary, to meet it in F.

Then, because AB is equal to AE or AG, CE=AC+AB, the sum of the sides; and CG-AC-AB, the difference of the sides. Also, because BD is equal to DF (Prop. VI., B. III.); when the perpendicular falls within the triangle, CF-CDDF-CD-DB, the difference of the segments of the base. But when the perpendicular falls without the triangle, CF= CD+DF=CD+DB, the sum of the segments of the base.

Now in either case, the rectangle ČE×CG is equivalent to CBXCF (Prop. XXVIII., Cor. 2); that is,

(AC+AB) × (AC-AB)=(CD+DB) × (CD-DB). Therefore, if from any angle, &c.

Cor. If we reduce the preceding equation to a proportion (Prop. II., B. II.), we shall have

BC: AC+AB:: AC-AB: CD-DB;

that is, the base of any triangle is to the sum of the two other sides, as the difference of the latter is to the difference of the segments of the base made by the perpendicular.

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The diagonal and side of a square have no common measure.

Let ABCD be a square, and AC its diagonal; AC and AB have no common

measure.

D

E

F

In order to find the common measure, if there is one, we must apply CB to CA as often as it is contained in it. For this purpose, from the center C, with a radius CB, describe the semicircle EBF. We A G B perceive that CB is contained once in AC, with a remainder AE, which remainder must be compared with BC or its equal AB.

Now, since the angle ABC is a right angle, AB is a tangent to ne circumference; and AE: AB:: AB: AF ( rop.

D

E

XXVIII., Cor. 1). Instead, therefore, of comparing AE with AB, we may substitute the equal ratio of AB to AF. But AB is contained twice in AF, with a remainder AE, which must be again compared with AB. Instead, however, of comparing AE with AB, we may again employ the equal ratio of AB to AF. A Hence at each operation we are obliged to compare AB with AF, which leaves a remainder AE; from which we see that the process will never terminate, and therefore there is no common measure between the diagonal and side of a square that is, there is no line which is contained an exact number of times in each of them.

G B

BOOK V.

PROBLEMS.

Postulates.

1. A straight line may be drawn from any one point to any other point.

2. A terminated straight line may be produced to any length in a straight line.

3. From the greater of two straight lines, a part may be cut off equal to the less.

4. A circumference may be described from any center, and with any radius.

PROBLEM I.

To bisect a given straight line.

Let AB be the given straight line which it is required to bisect.

From the center A, with a radius great

er than the half of AB, describe an arc of A

a circle (Postulate 4); and from the center B, with the same radius, describe another arc intersecting the former in D and

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E. Through the points of intersection, draw the straight line DE (Post. 1); it will bisect AB in C.

For, the two points D and E, being each equally distant from the extremities A and B, must both lie in the perpendicular, raised from the middle point of AB (Prop. XVIII., Cor., B. I.). Therefore the line DE divides the line AB into two equal parts at the point C.

PROBLEM II.

To draw a perpendicular to a straight line, from a given point in that line.

Let BC be the given straight line, and A the point given in it; it is required to draw a straight line perpendicular to BC through the given point A.

B

A

In the straight line BC take any point B, and make AC equal to AB (Post. 3). From B as a center, with a radius greater than BA, describe an arc of a circle (Post. 4); and from C as a center, with the same radius, describe another arc intersecting the former in D. Draw AD (Post. 1), and it will be the perpendicular required.

For, the points A and D, being equally distant from B and C, must be in a line perpendicular to the middle of BC (Prop. XVIII., Cor., B. I.). Therefore AD has been drawn perpendicular to BC from the point A.

Scholium. The same construction serves to make a right angle BAD at a given point A, on a given line BC.

PROBLEM III.

To draw a perpendicular to a straight line, from a given point without it.

Let BD be a straight line of unlimited length, and let A be a given point without it. It is required to draw a perpendicular to BD from the point A.

Take any point E upon the other side of BD; and from the center A, with the B radius AE, describe the arc BD cutting the line BCD in the two points B and D. From the points B and D as centers, de

A

scribe two arcs, as in Prob. II., cutting each other in F. Join AF, and it will be the perpendicular required.

For the two points A and Fare each equally distant from the points B and D; therefore the line AF has been drawn perpendicular to BD (Prop. XVIII., Cor., B. I.), from the given point A.

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