THEOREM.

PROPOSITION XII.

The angle formed within a circle, by the intersection of two chords, is measured by half the sum of the two intercepted arcs.

Let the two chords AB, CD intersect at the point F; then will the angle AFC, or its equal BFD, be measured by half the sum of the arcs AC and BD.

Draw the chord AD; then will the angle AFC be equal to the sum of the

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two angles FAD, FDA, (B. I, Prop. xxv.) The angle FAD is measured by half the arc DB, (B III, Prop. vIII;) for the same reason, the angle FDA is measured by half the arc AC. Therefore the angle AFC, which is equal to the sum of the two angles FAD, FDA, is measured by half the sum of the two arcs DB and AC.

In the same way, by drawing the chord BD, it may be shown that the angle AFD, or its equal BFC, is measured by half the sum of the two arcs AD and BC.

PROPOSITION XIII.

THEOREM. The angle formed, out of a circle, by two secants, is measured by half the difference of the intercepted arcs.

Let the angle F be formed by the two secants FAB, FCD; this angle will be measured by half the difference of the two arcs BD and AC intercepted by the two secants.

Draw the chord AG parallel to FD. Then, because the lines FD and AG are parallel, and BF meets them, the exterior angle BAG is equal to the interior and opposite angle BFD, (B. I, Prop. xx.) But the angle BAG, being at the circumference, is measured by half the

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arc BG, (B. III, Prop. viii,) or by half the difference of the arcs BD and GD; therefore the angle BFD is also measured by half the difference of the arcs BD and GD.

Again, because the chords CD and AG are parallel, the arc AC is equal to GD, (B. III, Prop. x ;) therefore the difference of the two arcs BD and AC is equal to the difference of the two BD and GD, and consequently the angle BFD is measured by half the difference of the arcs BD and AC.

THEOREM.

PROPOSITION XIV.

The angle formed by two tangents is measured by half the difference of the two intercepted arcs.

Let FB, FD be two tangents to a circumference at the points A, C; then will the angle F be measured by half the difference of the two arcs AGC, CHA.

Draw the chord AG parallel to FD. Then, because the lines AG and FD are parallel, and BF meets them, the exterior angle

BAG is equal to the interior and opposite angle BFD, (B. I, Prop. xx.) But the angle BAG, formed by a tangent and chord, is measured by half the arc AKG, (B. III, Prop. vi;) therefore the equal angle F will also be measured by half the same arc AKG, or half the difference of the arcs AGC and GC, or half the difference of the two, AGC and CHA, (B. III, Prop. x1.)

Cor. In like manner, it is proved that the angle F, formed by a tangent FCD and a secant FAB, is measured by half the difference of the two intercepted arcs BGC and CA.

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PROPOSITION XV.

PROBLEM. On a given line as a chord, to describe a segment of a circle, capable of containing an angle of a given magnitude.

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Let AB be the given line, and K the given angle. Draw CD, bisecting AB at right-angles; also, at B, draw BF perpendicular to AB; then draw BG, making the angle FBG equal to the given angle K, (B. I, Prop. x.)

Joining GA, and comparing the two triangles GCB and GCA, we have GC common, and the side CB equal to CA; also the angle GCB equal to the angle GCA, each being a right-angle; therefore GB is equal to GA, (B. I, Prop. III.) Hence, if, with G as a centre, a circumference be described, passing through the point B, it will also pass through the point A, and the segment AHB will be the one required.

Since the angle AGB at the centre is measured by the arc ALB, and the angle AHB at the circumference is measured by half the same arc, (B. III, Prop. vII,) it follows that the angle AHB is half the angle AGB; therefore the angle AHB, inscribed in the segment AHB, is equal to the angle BGC, the half of BGA; but the angle BGC is equal to the alternate angle FBG, (B. I, Prop. XVII,) which was made equal to the given angle K; therefore the angle AHB, in the segment AHB, is equal to the given angle K.

Schol. If an angle ALB be inscribed in the other segment of the circle, it will, with the angle AHB, be equal to the sum of two right-angles, (B. III, Prop. vii, Cor. 4.) Hence, when the given angle K is a right-angle, the line AB will be a diameter; and when the angle K is acute, then will the angle ALB be obtuse.

When the given angle K is obtuse, if we find a segment capable of containing the angle which remains when K is taken from two right-angles, then will the opposite segment contain angles equal to the given angle K.

PROPOSITION XVI.

PROBLEM. Through a given point, to draw a tangent to a given circumference.

If the given point A is situated in the circumference, draw the radius CA; and through the point A draw BD perpendicular to CA, and it will be the tangent required, (B. III, Prop. v.)

If the point A is situated. without the circumference, join AC, bisect it at F, and with F as a centre, and a radius equal to FA or FC, describe a circumference cutting the given circumference at the points B and D. Draw AB and AD, and they will each be tangents as required. For, joining CB and CD, we know that the angles ABC,

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ADC are right-angles, being inscribed in a semicircle, (B. III, Prop. VII, Cor. 3;) hence the lines AB, AD are tangent to the given circumference, (B. III, Prop. v.)

In this case, the two tangents are of the same length. For, in the two right-angled triangles ABC, ADC, we. have AC a common hypothenuse, and the sides BC, CD equal, each being a radius of the same circle; therefore those triangles are equal, (B. II, Prop. vIII, Cor. 2.) Consequently, AB is equal to AD.