53. Note. We may now see the reason for the construction given in § 28. We know that the triangle whose vertices are the points B, G, and H has its sides respectively equal to those of the triangle whose vertices are the points E, K, and L. Then, the triangles are equal by § 52, and the homologous angles B and E are equal.

54. If a perpendicular be erected at the middle point of a straight line,

I. Any point in the perpendicular is equally distant from the extremities of the line.

II. Any point without the perpendicular is unequally distant from the extremities of the line.

At the middle point D of any line AB erect a 1 DC. From E, any point in DC, draw lines to A and B. We now have:

I. Given line CDL to line AB at its middle point D, E any point in CD, and lines AE (a) and BE (b).

To Prove

a=b.

Proof. 1. In ▲ ADE and BDE, DE = DE.

[Two ▲ are equal when two sides and the included of one are equal respectively to two sides and the included

of the other.]

(§ 46)

[In equal figures, the homologous parts are equal.]

(§ 48)

Draw line AB. At D, the middle point of AB, draw DC L AB. From F, any point without CD, draw lines FA and FB. We then have :

II. Given line CD L to line AB at its middle point D, F any point without CD, and lines AF and BF.

To Prove AF and BF unequal.

Proof. 1. Let AF intersect CD at E, and draw line BE. 2. We have

BEEF BF.

[A str. line is the shortest line between two points.]

(Ax. 7)

[If a be erected at the middle point of a str. line, any point in the is equally distant from the extremities of the line.]

4. Substituting for BE its equal AE,

AE + EF> BF, or AF > BF.

(§ 54, I)

55. It follows from Prop. VI that every point which is equally distant from the extremities of a straight line, lies in the perpendicular erected at the middle point of the line.

56. A straight line is determined by any two of its points (§ 35); whence,

Two points, each equally distant from the extremities of a straight line, determine a perpendicular to that line at its middle point.

57. From the equal triangles ADE and BDE, in the figure of § 54, I,

LAED BED;

since homologous parts of equal figures are equal.

Then, if lines be 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 perpendicular.

58. We may now see the reasons for the constructions given in §§ 24, 25, 26, and 27.

In § 24, by construction, points C and F are each equally distant from points D and E; and CF is perpendicular to AB by § 56.

In § 25, points C and Fare each equally distant from points D and E; and in § 26, points C and D are each equally distant from points A and B.

In § 27, points O and E are each equally distant from points C and D; then, OE is perpendicular to straight line CD at its middle point (§ 56), and ▲ AOE = ▲ BOE by § 57.

Ex. 10. If in triangles ABD and BCD, sides AB, BD, and AD are equal, respectively, to sides CD, BD, and BC, what angles of the triangles are equal?

A

B

Ex. 11. Two lines of unequal length bisect each other at right angles. Show that any point in either line is equidistant from the extremities of the other. (§ 54.)

Ex. 12. If lines be drawn from the extremities of a straight line to any point in the perpendicular erected at its middle point, they make equal angles with the line. (Figure of § 54, I. Use § 48.)

ns!

PROP. VII. THEOREM

as 59. From a given point without a straight line, but one perpenAdicular can be drawn to the line.

(It follows from § 25 that, from a given point without a straight line, a perpendicular can be drawn to the line.)

np

-dis

p

From point C without line AB, draw CD 1 AB (§ 25). We then have :
Given point C without line AB, and line CD ↓ AB.

To Prove CD the only which can be drawn from C to AB. Proof. 1. If possible, let CE be another from C to AB. 2. Extend CD to C', making C'D = CD, and draw line EC'; represent CED by a, and Z C'ED by b.

3. Since ED is 1 CC at its middle point D, La = Zb.

[If lines be drawn to the extremities of a str. line from any point in the erected at its middle point, they make equal ▲ with the L.]

(§ 57)

4. By hyp., a is a rt. Z.

5. Then, b is a rt. ≤, and Za+b=two rt. 4.

6. Then line CEC is a str. line.

[If the sum of two adj. is equal to two rt. 4, their ext. sides lie in the same str. line.]

(§ 20)

7. This is impossible, for, by cons., CDC is a str. line.
[But one str. line can be drawn between two points.]

(Ax 6.)