(Prop. XVI., B. III.); and the angle C is measured by half the same arc, therefore the angle ABD is equal to C, and the two triangles ABD, ABC are equiangular, and, consequently, similar; therefore (Prop. XVIII.)

AC AB: AB: AD;

Therefore, if from a point, &c.

Cor. 1. If from a point without a circle, a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external segment.

Cor. 2. If from a point without a circle, two secants be drawn, the rectangles contained by the whole secants and their external segments will be equivalent to each other; for each of these rectangles is equivalent to the square of the tangent from the same point.

Cor. 3. If from a point without a circle, two secants be drawn, the whole secants will be reciprocally proportional to their external segments.

If an angle of a triangle be bisected by a line which cuts the base, the rectangle contained by the sides of the triangle, is equivalent to the rectangle contained by the segments of the base, together with the square of the bisecting line.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; the rectangle BAXAC is equivalent to BDXDC together with the square of AD.

C

D

E

Describe the circle ACEB about the triangle, and produce AD to meet 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 AEC, for they are in the same segment (Prop. XV., Cor. 1, B. III), the triangles ABD, AEC are mutually equiangular and similar; therefore (Prop. XVIII.)

BA: AD: EA: AC; consequently (Prop. I., B. II.),

BAXAC AD×AE.

But AE=AD+DE; and multiplying each of these equals by AD, we have (Prop. III.) ÁĎ×ÅE=AD2+AD ×DE. But ADXDE=BDx DC (Prop. XXVII.); hence

BAX AC-BDx DC+AD'.

Therefore, if an angle, &c.

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 in- B scribed in a circle, and let the diagonals AC, BD be drawn; the rectangle ACX BD is equivalent to the sum of the two rectangles AD× BC and AB×CD.

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

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, ADX 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,

AB: AE:: BD: CD;

ABXCD BDX AE.

=

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

BD×

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) × (CD−DB). From A as a center, with a radius equal to AB, the short

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 CEXCG 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.

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 he circumference; and AE: AB :: AB : AF (. rop.

D

XXVIII., Cor. 1). Instead, therefore, of
comparing AE with AB, we may substi-
tute the equal ratio of AB to AF. But
AB is contained twice in AF, with a re-
mainder AE, which must be again com-
pared with AB. Instead, however, of E
comparing AE with AB, we may again

F

employ the equal ratio of AB to AF. A G B 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.

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 cen-
ter B, with the same radius, describe an-
other arc intersecting the former in D and

-B

E. Through the points of intersection, draw the straight line DE (Post. I); 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.