Let to bisect the finite straight line AB Let the equilibriu

Let to bisect the finite straight line AB. Let the equilibrium triangle ABC be constructed on it [prop.1.1], and let the angle ACB be bisected by the straight line CD [prop.1.9]; I say that the straight line AB has been bisected at the point D. For, Since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively and the angle ACD is equal to the angle BCD; therefore the base AD this equal to the base BD [SAS]. Therefore the given finite straight line AB has been bisected at D, Note that Euclid bisects the angle with a line that intersects AB at a point D that lies between A and B. While it is clear that the angle bisector must contain an interior point of the triangle, details are missing that justify placing the point D where Euclid has it. Let CE vector be the ray that bisects angleACB where E is a point interior to the triangle angleABC. Prove that ray CE vector intersects angle ABC at a point D on AB bar and that D must lie strictly between A and B. Proposition 1.10 gives us not only the midpoint of the given line segment but constructs the perpendicular bisector of that segment. In his next two propositions, Euclid explains how to draw a line through a point C and perpendicular to a given line AB. If C lies on AB, this is called erecting a perpendicular, while if C does not lie on the line the procedure is called dropping a perpendicular. Given a line and a point on the line, one can construct a line through the given point and perpendicular to the given line.

Solution

The fact that E is an interior point of ABC

means that:

(1) E is on the same side of BC as A ;

(2) E is on the same side of AC as B ; and

(3) E is on the same side of AB as C(although we don\'t need this).

First we prove that CE cannot be parallel to AB

. Assume the contrary.

Note that A and B are on the same side of line m=CE. Denote by H the half-plane to which they belong. Select a point F on line CE but on the other side of C from E. By (1), F and A are on opposite sides of BC (since E and F are). Thus the segment AF intersects line CB. But since segment AF is situated in H, the point of intersection G must in fact be on the ray CB. Since G is between A and F, points F and G are on the same side of line AC. But since G is on ray CB, points G and B are also on the same side of AC. By transitivity, points B and F are on the same side of AC. By the definition of F, this shows that E and B are on opposite sides of AC, contradicting (2). Therefore CE and AB are not parallel.

It follows that line CE intersects line AB at some point D. We wish to prove that D is between A and B. We do this by eliminating the remaining possibilities. D cannot coincide with A, because then E would lie on the line CA, contradicting (2). Similarly, D cannot be B. We will now show that A cannot be between B and D. Since it can be shown in the same way B cannot be between A and D , this is the last case to be addressed.

Assume therefore that A is between B and D. Thus B and D are on opposite sides of CA. By (2) therefore, E and D are on opposite sides of CA, hence of C. On the other hand, since A and D are on the same side of B, they\'re on the same side of CB. By (1) therefore D and E are on the same side of CB, hence of C . This is a contradiction.

From the foregoing we conclude that D is between A and B. It remains only to show that D is on the ray CE. Since D and B are on the same side of A, they are on the same side of CA. Therefore, by (2), E and D are on the same side of CA, hence of C. Thus D is on the ray CE.