Cor. 1.—Every triangle must have at least two acute angles. Cor. 2.—If two angles of a triangle be unequal, the lesser must be acute. Exercise. Prove Prop. xvii. without producing a side. PROP. XVIII.—Theorem. If in any triangle (ABC) one side (AC) be greater than another (AB), the angle opposite to the greater side is grater than the angle opposite to the less. Dem.—From AC cut off AD equal to AB [iii]. Join BD (Post. i.). Now since AB is equal to AD, the triangle ABD is isosceles; therefore [v.] the angle ADB is equal to ABD; but the angle ADB is greater than the angle ACB [xvi.]; therefore ABD is greater than ACB. Much more is the angle ABC greater than the angle ACB. Or thus: From A as centre, with the lesser side AB as radius, describe the circle BED, cutting BC in E. Join AE. Now since AB is equal to AE, the angle AEB is equal to ABE; but AEB is greater than ACB (xvi.); therefore ABE is greater than ACB. Exercises. 1. If in the second method the circle cut the line CB produced through B, prove the Proposition. 2. This Proposition may be proved by producing the less side. 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. 2. A line from the vertex of an isosceles triangle to any point in the base is less than either of the equal sides, but greater if the point be in the base produced. 3. Three equal lines could not be drawn from the same point to the same line. 4. The perpendicular is the least line which can be drawn from a given point to a given line; and of all others that may be drawn to it, that which is nearest to the perpendicular is less than any one more remote. 5. If in the fig., Prop. xvi., AB be the greatest side of the △ ABC, BF is the greatest side of the △ FBC, and the angle BFC is less than half the angle ABC.