3. If two of the opposite sides of a quadrilateral be respectively the greatest and least, the angles adjacent to the least are greater than their opposite angles.
4. In any triangle, the perpendicular from the vertex opposite the side which is not less than either of the remaining sides falls within the triangle.
PROP. XIX.—Theorem.
If one angle (B) of a triangle (ABC) be greater than another angle (C), the side (AC) which it opposite to the greater angle is greater than the side (AB) which is opposite to the less.
Dem.—If AC be not greater than AB, it must be either equal to it or less than it. Let us examine each case:—
1. If AC were equal to AB, the triangle ACB would be isosceles, and then the angle B would be equal to C [v.]; but it is not by hypothesis; therefore AB is not equal to AC.
2. If AC were less than AB, the angle B would be less than the angle C [xviii.]; but it is not by hypothesis; therefore AC is not less than AB; and since AC is neither equal to AB nor less than it, it must be greater.
Exercises.
1. Prove this Proposition by a direct demonstration.