2d. Let the centre of the circle fall within the inscribed

angle BAC; then the measure of the angle BAC is one-half of the arc BC.

For, draw the diameter AD. The measure of the angle BAD is, by the first case, one-half the arc BD; and the measure of the angle CAD is one-half the arc CD; therefore the measure of the sum of the angles BAD and CAD is one-half the sum

of the arcs BD and CD; that is, the measure of the angle BAC is one-half the arc BC.

3d. Let the centre of the circle fall without the inscribed angle BAC; then the measure of the angle BAC is one-half the arc BC.

D

For, draw the diameter AD. The measure of the angle BAD is, by the first case, one-half the arc BD; and the measure of the angle CAD is one-half the arc CD; therefore the measure of the difference of the angles BAD and CAD is one-half the difference of the arcs BD and CD; that is, the measure of the angle BAC is one-half the arc BC.

49. COROLLARY. An angle inscribed in a semicircle is a right angle.

C D

EXERCISE.

Theorem.-The opposite angles of an inscribed quadrilateral

are supplements of each other.

PROPOSITION XV.-THEOREM.

50. An angle formed by a tangent and a chord is measured by one-half the intercepted arc.

Let the angle BAC be formed by the tangent AB and the chord AC; then it is measured by one-half the intercepted arc AMC.

For, draw the diameter AD. The angle BAD, being a right angle (Proposition IX.), is measured by one-half

M

B

B

N

the semi-circumference AMD; and the angle CAD is measured by one-half the arc CD; therefore the angle BAC, which is the difference of the angles BAD and CAD, is measured by one-half the difference of AMD and CD; that is, by one-half the arc AMC.

Also, the angle B'AC is measured by one-half the intercepted arc ANC. For, it is the sum of the right angle B'AD and the angle CAD, and is measured by one-half the sum of the semi-circumference AND and the arc CD; that is, by one-half the arc ANC.

B

E

EXERCISE.

Prove Proposition XV. by the aid of this figure, OE being a radius perpendic ular to AC.

Suggestion. Complements of the same angle are equal.

Di

PROPOSITION XVI.-THEOREM.

51. An angle formed by two chords, intersecting within the circumference, is measured by one-half the sum of the arcs intercepted between its sides and between the sides of its vertical angle. Let the angle AEC be formed by the chords AB, CD, intersecting within the circumference; then will it be measured by one-half the sum of the arcs AC and BD, intercepted between the sides of AEC and the sides of its vertical angle BED.

For, join AD. The angle AEC is equal

B

E

D

to the sum of the angles EDA and EAD, and these angles are measured by one-half of AC and one-half of BD, respectively; therefore the angle AEC is measured by one-half the sum of the arcs AC and BD.

EXERCISE.

Prove Proposition XVI. by the aid of this figure, DF being drawn parallel to AB. (v. Proposition XI.)

PROPOSITION XVII.-THEOREM.

52. An angle formed by two secants, intersecting without the circumference, is measured by one-half the difference of the intercepted arcs.

Let the angle BAC be formed by the secants AB and AC; then will it be measured by one-half the difference of the arcs BC and DE.

For, join CD. The angle BDC is equal to the sum of the angles DAC and ACD; therefore the angle A is equal to the differ

B

A

D

E

Ο

ence of the angles BDC and ACD. But these angles are measured by one-half of BC and one-half of DE respectively; hence the angle A is measured by one-half the difference of BC and DE.

EXERCISE.

Prove Proposition XVII. by the aid of Proposition XI., drawing a suitable figure.

PROPOSITION XVIII.-THEOREM.

53. An angle formed by a tangent and a secant is measured by half the difference of the intercepted arcs.

For the angle A is equal to BDC minus ABD, by I., Proposition XXVI., Corollary. 54. COROLLARY. An angle formed by two tangents is measured by half the difference of the intercepted arcs.

B

EXERCISE.

1. Prove Proposition XVIII. and its Corollary by the aid of Proposition XI.

2. Theorem.-If, through the point of contact of two tangent circles, two secants are drawn, the

chords joining the points where the secants cut the circles are parallel. Suggestion. FED CEG,

DBE

CAE.

Consider, also, the case where the given circles are internally tangent.

E

PROBLEMS OF CONSTRUCTION.

Heretofore our figures have been assumed to be constructed under certain conditions, although methods of constructing them have not been given. Indeed, the precise construction of the figures was not necessary, inasmuch as they were only required as aids in following the demonstration of principles. We now proceed, first, to apply these principles in the solution of the simple problems necessary for the construction of the plane figures already treated of, and then to apply these simple problems in the solution of more complex ones.

All the constructions of elementary geometry are effected solely by the straight line and the circumference, these being the only lines treated of in the elements; and these lines are practically drawn, or described, by the aid of the ruler and compasses, with the use of which the student is supposed to be familiar.

PROPOSITION XIX.-PROBLEM.

55. To bisect a given straight line.

Let AB be the given straight line.

With the points A and B as centres, and with a radius greater than the half of AB, describe arcs intersecting in the two points D and E. Through these points draw the straight line DE, which bisects AB at the point C. For, D and E being equally distant from A and B, the straight line DE is perpendicular to AB at its middle point (I., Proposition XVIII.).

B