75. It is often more convenient to express the magnitude of an angle in some other way than by stating how many right angles the given angle contains. To obtain another method, a right angle is divided into ninety equal parts, called degrees. The magnitude of an angle may, then, be expressed by stating how many degrees the given angle contains.

EXERCISES.

28. How many degrees in a straight angle? In all the angular magnitude about a point?

29. How many degrees in the supplement of two thirds of a right angle?

30. How many degrees in an angle whose complement equals one fourth of its supplement?

31. The supplement of ten degrees is how much more than the complement of ten degrees?

32. The supplement of any acute angle is how much more than the complement of the same acute angle? Note.- In elementary geometry it is only acute angles which have complements, but either acute or obtuse angles have supplements. 33. The supplement of the complement of any acute angle is how much more than the angle itself?

34. The complement of the supplement of an obtuse angle is how much less than the given obtuse angle?

35. Two angles are complements of each other, and the greater exceeds the less by 38 degrees. What are the angles?

36. If the bisectors of two adjacent angles are perpendicular to each other, the angles are supplements of each other.

PROPOSITION XIII. THEOREM.

76. From a point without a line, one, and only one, perpendicular can be drawn to the line.

G

A

D

P

N

Let C D represent the line, and A the point.

I. To prove that one perpendicular can be drawn from A to the line CD.

SUG. 1. Draw any line, as M N, and at some point of this line, as O, erect the L, ОР.

PROP. I.

SUG. 2. Place the line, MN, upon the line, CD, and move it back and forth in CD.

SUG. 3. Is it possible for O P to embrace the point A? Therefore

II. Only one perpendicular can be drawn to the

Let A O represent a perpendicular from the point, A, to

the line, C D.

To prove that no other perpendicular can be drawn from the point A to the line C D.

SUG. 1. If another

sented by A M.

can be drawn, let it be repre

SUG. 2. Extend AO to B, making O BA O, and connect M and B.

SUG. 3. In the AS A OM and B O M, compare A O with B O, MO with M O, and ZA O M with Z BOM. SUG. 4. Now, compare AMO with BM O. Give auth.

SUG. 5. Then, if by construction ZA MO is a rt. Z, is line A M B a straight or a broken line? Why?

SUG. 6. Then, how many straight lines are drawn from A to B?

SUG. 7. What, then, do you conclude about the statement that A M B is a straight line?

SUG. 8. Then, what do you conclude as to the possibility of AM being a from A to C D?

SUG. 9. Then, how many Is can be drawn from a point to a straight line?

Therefore

Ex. 37. Prove that the bisectors of two supplementary adjacent angles are perpendicular to each other.

Ex. 38. If A B and A C are equal sides of a triangle, and if B M and CN are bisectors of the angles B and C respectively, prove that the triangles A B M and ACN are equal, and also that the triangles BCN and CB M are equal.

Ex. 39. A line drawn through the vertex of an angle, perpendicular to the bisector of the angle, makes equa! angles with the sides of the given angle.

MODEL.

PROPOSITION XIII - PART II. THEOREM.

77. Only one perpendicular can be drawn from a point to a line.

Let A O represent a perpendicular from the point A to the line C D.

To prove that no other perpendicular can be drawn from the point A to the line CD.

If another

by A M.

can be drawn, let it be represented

Extend A O to B, so that A 0 = OB, and connect M and B.

=

In the As A O M and B OM, A O BO (by cons.), MOMO (identical) and ZA OM = LBOM (each being a rt. ).

PROP. IV

If two As bave two sides and the included of one, equal to two sides and the included of the other, each to each, the triangles are equal in all respects.

Therefore, the AS A O M and B O M are equal in all respects, and hence ZA MO LBMO.

Since LA MO is a rt. Z, by construction, and ZAMO equals BMO, AMB is a straight line. PROP. XII.

If two adjacent angles are together equal to two right angles, their exterior sides form a straight line.

But AO B is a straight line, hence A M B cannot be a straight line. Ax. 11.

Therefore, only one can be drawn from a point to a

line.

Ex. 40. If two vertical angles are bisected by two straight lines, prove that the bi

sectors together form one and the

same straight line. Prove that NNO M is a straight line.

SUG. See Prop. XII.

-M

Ex. 41. If A B C be an isosceles triangle, whose vertex is A, and whose base is B C, and if M be the middle point of A C, and N the middle point of A B, and if the lines B M and CN intersect at O, prove (1) that B M equals C N, (2) that the triangles B C M and C B N are equal, (3) that the triangles COM and B O N are equal, and (4) that the triangle B O C is isosceles.

Ex. 42. If A B C is an equilateral triangle, and D A and F are points in the sides A B, B C, and CA, espectively, such that A D B E = C F, prove that the triangle D E F is equilateral.

=

Ex. 43. If two lines, A B and C D, intersect in the point O, and if the lines A C and B D be drawn, ▲ R + CD is greater than A C + B D.

Ex. 44. If A B C and A B D are two triangles on the same base, and on the same side of it, such that A C equals B D, and A D equals B C, and A D and BC intersect at O, prove (1) that the triangles A B C and A D are equal in all respects, (2) that the triangles AOC and B O D are equal in all respects, and (3) that the triangle AO B is isosceles.