NOTE. This is the converse of Prop. 18. We have here a second instance of the use of the indirect method for proving a converse (see Prop. 6). It should be carefully remembered that this proposition is merely equivalent to the following :— 'If two angles of a triangle are unequal, the side which subtends the greater is greater than the side subtending the less.' The proposition includes the following 'If two angles of a triangle are unequal, the sides which subtend them are unequal.' This is the obverse of Prop. 6, and is an immediate inference from Prop. 5. Ex. 24. Of all straight lines that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest; and of the others those which make equal angles with the perpendicular are equal; and that which makes a greater angle with the perpendicular is greater than that which makes a less angle. PROPOSITION 20. THEOREM. Any two sides of a triangle are together greater than the third. Let ABC be a ▲, any two sides of it are together greater than the third side. A B Produce BA to D, making AD equal to AC; join CD. Then AD AC, .. LADC= 4 ACD. But BCD > LACD, ... 4 BCD > LADC. [I. 3. [CONST. [I. 5. [AX. 9. .. side BD of ▲ BCD opp. to BCD > side BC in same opp. 4 ADC. But BD=BA and AC, .'. BA, AC > BC. [I. 19. [CONST. In the same manner it may be shown that AB, BC > AC, and that BC, CA > BA. Ex. 25.-Any side of a triangle is greater than the difference of the other two sides. Prove this (1) independently, (2) as a deduction from Prop. 20. Ex. 26. Any three sides of a quadrilateral are together greater than the fourth. Ex. 27.-Any two sides of a triangle are together greater than twice the median line which bisects the base (see figure of Prop. 16). Ex. 28.-The sum of the distances of any point from the three vertices of a triangle is greater than half the sum of the sides. Ex. 29.—Enunciate and prove a similar theorem as regards a quadrilateral. Can it be extended to other polygons (figures with more sides than three)? The four sides of a quadrilateral are together greater than the diagonals. Are the five sides of a pentagon (five-sided figure) greater than its five diagonals? Ex. 30. Two triangles are constructed, each by joining the alternate vertices of a hexagon (six-sided figure). Show that the perimeter of the hexagon is greater than half the sum of the perimeters of the triangles. Ex. 31.-ABCDEF is a hexagon. Show that its perimeter is greater than two-thirds of the sum of the three diagonals AD, BE, CF. PROPOSITION 21. THEOREM. If from the ends of a side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. From the ends B, C of the side BC of the ▲ ABC let the two st. lines BD, CD be drawn to a point D within the A; BD, DC shall be less than BA, AC but 4 BDC shall be greater than BAC. ... BA, AC > BD, DC. Again ext. BDC > int. and opp. ▲ BEC. [DEMON. [I. 16. Also ext. 4 BEC > int. and opp. 4 BAC. ... 4 BDC > < BAC. P [I. 16. R Ex. 32. Straight lines, BP, PQ, QR, RC, are drawn forming a rectilineal path, BPQRC, from B to C but not crossing BC. A second rectilineal path, BSTVC, is drawn from B to C lying entirely between the first path and BC, and such that if any of the straight lines BS, ST, TV, VC which form it be produced either way, the produced parts will lie without the figure BSTVC. Show that the outside path is the longer. B (A generalisation of the first part of Prop. 21, to be proved in the same way.) Examine the effect of omitting the words in italics from the hypothesis. Ex. 33.-O is any point within the triangle ABC, show that OA, OB, OC are together less than AB, BC, CA together. |