Also, since the angle ACE is equal to the angle BCE, the arc AE, which measures the former (def. 57), is equal to the arc BE, which measures the latter, since equal angles must have equal measures.

Corol. Hence, a line bisecting any chord at right angles, passes through the centre of the circle.

THEOREM XLII.

If more than two equal lines can be drawn from any point within a circle to the circumference, that point will be the centre.

Let ABC be a circle, and D a point within it: then, if any three lines, DA, DB, DC, drawn from the point D to the circumference, be equal to each other, the point D will be the centre.

B

Draw the chords AB, BC, which let be bisected in the points E, F, and join DE, DF.

D

Then, the two triangles DAE, DBE, have the side DA equal to the side DB by supposition, and the side

AE equal to the side EB by hypothesis, also the side DE common: therefore, these two triangles' are identical, and have the angles at E equal to each other (th. 5); consequently, DE is perpendicular to the middle of the chord AB (def. 11), and therefore passes through the centre of the circle (corol. th. 41). In like manner, it may be shown that DF passes through the centre. Consequently, the point D is the centre of the circle, and the three equal lines DA, DB, DC, are radii. Q. E. D.

THEOREM XLIII.

If two circles, placed one within another, touch, the centres of the circles and the point of contact will be ali in the same right line.

Let the two circles ABC, ADE, touch one another internally in the point A; then will the point A and the centres of those circles be all in the same right line.

Let F be the centre of the circle ABC, through which draw the diameter AFC. Then, if the centre of the other circle can be out of this line AC, let it be supposed in some other point as G; through which draw the line FG, cutting the two circles in B and D.

B

F

D

G

E

Now, in the triangle AFG, the sum of the two sides FG, GA, is greater than the third side AF (th. 10), or greater than its equal radius FB. From each of these take away the common part FG, and the remainder GA will be greater than the remainder GB. But the point G being supposed the centre of the inner circle, its two radii, GA, GD, are equal to each other; consequently, GD will also be greater than GB. But ADE being the inner circle, GD is necessarily less than GB. So that GD is both greater and less than GB; which is absurd. To get quit of this absurdity we must abandon the supposition that produced it, which was that G might be out of the line AFC. Consequently, the centre G cannot be out of the line AFC. Q. E. D.

THEOREM XLIV.

If two circles touch one another externally, the centres of the circles and the point of contact will be all in the same right line.

Let the two circles ABC, ADE, touch one another externally at the point A; then will the point of contact A and the centres of the two circles be all in the same right line.

Let F be the centre of the circle ABC, through which draw the diameter AFC, and produce it to the other circle at E. Then, if the centre of the other circle ADE can be out of the line FE, let it, if possible, be supposed in some other point as G; and draw the lines AG, FBDG, cutting the two circles in B and D.

A

E

F

Then, in the triangle AFG, the sum of the two sides AF, AG, is greater than the third side FG (th. 10). But, F and G being the centres of the two circles, the two radii GA, GD, are equal, as are also the two radii AF, FB. Hence the sum of GA, AF, is equal to the sum of GD, BF; and, therefore, this latter sum also, GD, BF, is greater than GF, which is absurd. Consequently, the centre G cannot be out of the line EF. Q. E. D.

THEOREM XLV.

Any chords in a circle, which are equally distant from the centre, are equal to each other; or if they be equal to each other, they will be equally distant from the centre.

Let AB, CD, be any two chords at equal distances from the centre G; then will these two chords AB, CD, be equal to each other.

Draw the two radii GA, GC, and the two perpendiculars GE, GF, which are the equal distances from the centre G. Then, the two right-angled triangles, GAE, GCF, having the side GA equal the side GC, and the

A

side GE equal the side GF, and the angle at E equal to the angle at F, therefore those two triangles are identical (cor. 2, th. 34), and have the line AE equal to the line CF. But AB is the double of AE, and CD is the double of CF (th. 41); therefore AB is equal to CD (by ax. 6). Q. E. D.

Again, if the chord AB be equal to the chord CD; then will their distances from the centre, GE, GF, also be equal to each other.

For, since AB is equal CD by supposition, the half AE is equal the half CF. Also, the radii GA, GC, being equal, as well as the right angles E and F, therefore the third sides are equal (cor. 2, th. 34), or the distance GE equal the distance GF. Q. E. D.

A line perpendicular to the extremity of a radius, is a tangent to the circle.

Let the line ADB be perpendicular to the radius CD of a circle; then shall AB touch the circle in the point D only.

From any other point E in the line AB draw CFE to the centre, cutting the circle in F.

Then, because the angle D, of the triangle CDE, is a right angle, the angle at E is acute (cor. 3, th. 17), and consequently less than the angle D. But the greater side

A DEB

is always opposite to the greater angle (th. 9); therefore the side CE is greater

than the side CD, or greater than its equal CF.

Hence the point E is without the circle; and the same for every other point in the line AB. Consequently the whole line is without the circle, and meets it in the point D only.

THEOREM XLVII.

When a line is a tangent to a circle, a radius drawn to the point of contact is perpendicular to the tangent.

Let the line AB touch the circumference of a circle at the point D; then will the radius CD be perpendicular to the tangent AB. [See the last figure.] For, the line AB being wholly without the circumference except at the point D, every other line, as CE, drawn from the centre C to the line AB, must pass out of the circle to arrive at this line. The line CD is therefore the shortest that can be drawn from the point C to the line AB, and consequently (th. 21,) it is perpendicular to that line.

Corol. Hence, conversely, a line drawn perpendicular to a tangent, at the point of contact, passes through the centre of the circle.

The angle formed by a tangent and chord is measured by half the arc of that chord.

Let AB be a tangent to a circle, and CD a chord drawn from the point of contact C; then is the angle BCD measured by half the arc CDF, and the angle ACD measured by half the arc CGD.

Draw the radius EC to the point of contact, and the radius EF perpendicular to the chord at H.

Then the radius EF, being perpendicular to the chord CD, bisects the arc CFD (th. 41). Therefore CF is half

the arc CFD.

In the triangle CEH, the angle H being a right one, the sum of the two remaining angles E and C is equal to a right angle (cor. 3, th. 17), which is equal to the angle BCE, because the radius CE is perpendicular to the tan

A

B

gent. From each of these equals take away the common part or angle C, and there remains the angle E equal to the angle BCD. But the angle E is measured by the arc CF (def. 57), which is the half of CFD; therefore the equal angle BCD must also have the same measure, namely, half the arc CFD of the chord CD.

Again, the line GEF, being perpendicular to the chord CD, bisects the arc CGD (th. 41). Therefore CG is half the arc CGD. Now, since the line CE, meeting FG, makes the sum of the two angles at E equal to two right angles (th. 6), and the line CD makes with AB the sum of the two angles at C equal to two right angles; if from these two equal sums there be taken away the parts or angles CEH and BCH, which have been proved equal, there remains the angle CEG equal to the angle ACH. But the former of these, CEG, being an angle at the centre, is measured by the arc CG (def. 57); consequently the equal angle ACD must also have the same measure CG, which is half the arc CGD of the chord CD. Q. E. D.

Corol. 1. The sum of the two right angles is measured by half the circumference. For the two angles BCD, ACD, which make up two right angles,

are

measured by the arcs CF, CG, which make up half the circumference, FG being a diameter.

Corol. 2. Hence also one right angle must have for its measure a quarter of the circumference, or 90 degrees.

THEOREM XLIX.

An angle at the circumference of a circle is measured by half the arc that subtends it.

Let BAC be an angle at the circumference; it has for its measure, half the arc BC which subtends it.

D

E

B

For, suppose the tangent DE to pass through the point of contact A: then, the angle DAC being measured by half the arc ABC, and the angle DAB by half the arc AB (th. 48); it follows, by equal subtraction, that the difference, or angle BAC, must be measured by half the arc BC, which it stands upon. Q. E. D.

All angles in the same segment of a circle, or standing on the same arc, are equal to each other.

Let C and D be two angles in the same segment ACDB, or, which is the same thing, standing on the supplemental arc AEB; then will the angle C be equal to the angle D.

For, each of these angles is measured by half the arc AEB;

and thus, having equal measures, they are equal to

each other (ax. 11).

C

An angle at the centre of a circle is double the angle at the circumference, when both stand on the same arc.

Let C be an angle at the centre C, and D an angle at the circumference, both standing on the same arc or same chord AB; then will the angle C be double of the angle D, or the angle D equal to half the angle C.

For, the angle at the centre C is measured by the whole

arc AEB (def. 57), and the angle at the circumference D

B

F

is measured by half the same arc AEB (th. 49); therefore the angle D is only half the angle C, or the angle C double the angle D.

THEOREM LII.

An angle in a semicircle, is a right angle.

If ABC or ADC be a semicircle; then any angle D in that semicircle, is a right angle.

For, the angle D, at the circumference, is measured by half the arc ABC (th. 49), that is, by a quadrant of the circumference. But a quadrant is the measure of a right angle (cor. 4, th. 6; or cor. 2, th. 48). Therefore the angle D is a right angle.

B

THEOREM LIII.

The angle formed by a tangent to a circle, and a chord drawn from the point of contact, is equal to the angle in the alternate segment.

If AB be a tangent, and AC a chord, and D any angle in the alternate segment ADC; then will the angle D be equal to the angle BAC made by the tangent and chord of the arc AEC.

D

A

B

E

For, the angle D, at the circumference, is measured by half the arc AEC (th. 49); and the angle BAC, made by the tangent and chord, is also measured by the same half arc AEC (th. 48); therefore these two angles are equal (ax. 11).

THEOREM LIV.

The sum of any two opposite angles of a quadrangle inscribed in a circle, is equal to two right angles.

Let ABCD be any quadrilateral inscribed in a circle;

then shall the sum of the two opposite angles A and C, or

B and D, be equal to two right angles.

For the angle A is measured by half the arc DCB, which it stands upon, and the angle C by half the arc DAB (th. 49); therefore the sum of the two angles A and C is measured by half the sum of these two arcs, that is, by half

the circumference. But half the circumference is the measure of two right angles (cor. 4, th. 6); therefore the sum of the two opposite angles A and C is equal to two right angles. In like manner it is shown, that the sum of the other two opposite angles, D and B, is equal to two right angles. Q. E. D.

If any side of a quadrangle, inscribed in a circle, be produced out, the outward angle will be equal to the inward opposite angle.

If the side AB, of the quadrilateral ABCD, inscribed in a circle, be produced to E; the outward angle DAE will be equal to the inward opposite angle C.

C

For, the sum of the two adjacent angles DAE and DAB is equal to two right angles (th. 6); and the sum of the two opposite angles C and DAB is also equal to

two right angles (th. 54); therefore the former sum, of the two angles DAE and DAB, is equal to the latter sum, of the two C and DAB (ax. 1). From each of these equals taking away the common angle DAB, there remains the angle DAE equal the angle C. Q. E. D.

THEOREM LVI.

Any two parallel chords intercept equal arcs.

Let the two chords AB, CD, be parallel: then will the arcs AC, BD, be equal; or AC = BD.

Draw the line BC. Then, because the lines AB, CD, are parallel, the alternate angles B and C are equal (th. 12). But the angle at the circumference B, is measured by half the arc AC (th. 49); and the other equal angle

B