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.
6. If ABC be a △ having AB not greater than AC, a line AG, drawn from A to any point G in BC, is less than AC. For the angle ACB [xviii.] is not greater than ABC; but AGC [xvi.] is greater than ABC; therefore AGC is greater than ACG. Hence AC is greater than AG.
PROP. XX.—Theorem.
The sum of any two sides (BA, AC) of a triangle (ABC) is greater than the third.
Dem.—Produce BA to D (Post. ii.), and make AD equal to AC [iii.]. Join CD. Then because AD is equal to AC, the angle ACD is equal to ADC (v.); therefore the angle BCD is greater than the angle BDC; hence the side BD opposite to the greater angle is greater than BC opposite to the less [xix.]. Again, since AC is equal to AD, adding BA to both, we have the sum of the sides BA, AC equal to BD. Therefore the sum of BA, AC is greater than BC.
Or thus: Bisect the angle BAC by AE [ix.] Then the angle BEA is greater than EAC; but EAC = EAB (const.); therefore the angle BEA is greater than EAB. Hence AB is greater than BE [xix.]. In like manner AC is greater than EC. Therefore the sum of BA, AC is greater than BC.
Exercises.