II. Let C D be perpendicular to A B at its middle point D, and let F be any point without C D.

To prove that F is unequally distant from A and B.

Join A F and BF; let A F intersect C D at E, and draw B E. Since the straight line B F is shorter than the broken line BEF (a straight line being the shortest distance between two points),

BF is less than B E + E F.

Now we have already proved that B E A E, (the first part of this theorem.)

Hence B F is less than A E + E F, or B F is less than A F.
Therefore F is unequally distant from A and B.

43. Corollary 1. When the figure B D E is superposed upon A D E, the angles EBD and BED coincide with the angles E A D and A E D, respectively: that is,

angle E A D = angle E B D, and angle A E D = angle B E D. Hence, if straight lines are drawn to the extremities of a straight line from any point in the perpendicular erected at its middle point, they make equal angles with the line and with the perpendicular.

44. Corollary II. If a series of points, all of which satisfy a given condition, lie in a certain line, that line is called the locus of the points. For example, all the points which satisfy the condition of being equally distant from the extremities of a straight line lie in the perpendicular erected at its middle point.

In other words, the perpendicular at the middle point of a straight line is the locus of points which are equally distant from the extremities of the line.

45. Scholium. We know by axiom that a straight line is determined by any two of its points. It therefore follows that two points equally distant from the extremities of a line determine its perpendicular at its middle point.

46.

THEOREM VIII.

The perpendicular is the shortest line which can be

drawn from a point to a straight line.

Let CD be the perpendicular from C to A B, and C E any oblique line.

To prove that C D is less than C E.

Produce C D to C', making C' D = C D, and join E C. Then since E D is perpendicular to C C' at its middle point, CECE, (Theorem VII.)

But the straight line C C' being shorter than the broken line CE C', we have

CD CD is less than C E+ CE

Therefore, 2 C D is less than 2 C E, or C D is less than C E.

THEOREM IX.

47. From a given point outside a straight line, but one perpendicular can be drawn to that line.

Let C be the given point outside the line A B, and draw CD perpendicular to A B.

To prove that CD is the only perpendicular which can be drawn.

If possible, let C E be another perpendicular from C to A B. Produce C D to C', making C' D = C D, and then join C' E. Then since E D is perpendicular to C C' at its middle point, angle C'ED = angle C E D (§ 43.)

But C E D is a right angle (by hypothesis.)

Hence CE D is also a right angle, and the sum of the adjacent angles CED and C' E D is equal to two right angles, and therefore CE C' is a straight line (Theorem II.)

Now the latter is impossible, since but one straight line can be drawn between two given points (Axiom.)

Hence C E cannot be perpendicular to A B, and C D is the only perpendicular which can be drawn.

THEOREM X.

48. If two lines are drawn from a point to the extremities of a straight line, their sum is greater than the sum of two other lines similarly drawn, but enveloped by them.

Let A B and A C be drawn from the point A to the extremities of the line B C, and let D B and D C be two lines similarly drawn but enveloped by A B and A C.

To prove that AB+ A C is greater than D B + DC:
Produce B D to meet A C at E.

Then since the broken line B A E is greater than the straight

line B E, the line B A C is greater than B E C.

And since the broken line D E C is greater than the straight line D C, the line B E C is greater than B D C.

Therefore, A B+ A C is greater than D B + DC.

THEOREM XI.

49. If oblique lines are drawn from a point to a straight line.

I. Two oblique lines cutting off equal distances from the foot of the perpendicular are equal.

II. Of two oblique lines cutting off unequal distances from the foot of the perpendicular, the more remote is the greater.

A

I. Let the oblique lines C E and C F meet the line A B at equal distances from the foot of the perpendicular C D.

Since CD is perpendicular to E F at its middle point D, CECF (Theorem VII.)

II. Let the oblique lines C F and C G meet the line A B at unequal distances from the foot of the perpendicular C D, and let C G be the more remote.

To prove that C G is greater than C F:

Produce C D to C', making C' D = C D, and then join C' F and C' G.

C G. (Theorem VII).

is greater than the broken line

Now C F C F, and C G But the broken line C G C' CF C', which is enveloped by it. (Theorem X). Hence C G C G is greater than C F + C' F. Therefore 2 CG is greater than 2 C F, or C G is greater than C F.

50. Corollary I. Conversely two equal oblique lines from a point to a straight line, cut off equal distances from the foot of the perpendicular.

51.

Corollary II. Only two equal straight lines can be drawn from the same point to the same straight line.

THEOREM XII.

52. Two straight lines perpendicular to the same straight line are parallel.

Let the lines A B and C D be perpendicular to A C.

To prove that they are parallel:

If they are not parallel, they will meet if sufficiently produced (definition), and we should then have two perpendiculars from the same point to the same straight line, which is impossible. (Theorem IX).

Therefore, they cannot meet and are parallel.

THEOREM XIII,

53. A straight line perpendicular to one of two parallels is perpendicular to the other.

Let A B and CD be parallel lines, and let A C be perpendicular to A B.

To prove that A C is also perpendicular to C D:

Let C E be perpendicular to A C.

Then CE is parallel to A B. (Theorem XII).

But C D is parallel to A B (by hypothesis).

Hence CE must coincide with C D, since but one straight line can be drawn through the point C parallel to A B (by axiom). Therefore A C is perpendicular to C D.

54.

THEOREM XIV.

Two parallel lines are everywhere equally distant.

A