6. The altitude of a triangle is the perpendicular drawn from the vertex to the base, or the base produced; the altitude of a parallelogram is the perpendicular drawn from any point in one of its sides to the opposite side, or that side produced. 7. A straight line is said to be cut in extreme and mean ratio when the whole line is to one segment as that segment is to the other. As in the case of medial section, a straight line might be cut in extreme and mean ratio both internally and externally; but internal division only is generally implied by the phrase. PROPOSITION 1. THEOREM. Triangles and parallelograms of the same altitude are to one another as their bases. Let As ABC, ACD, and ||ms EC, CF have the same altitude, namely, the perpendicular drawn from A to BD, Produce BD both ways, and take any number of straight lines BG, GH, HK each = BC, and DL, LM, any number of them, each = CD; and join A with the points K, H, G, L, M. I. 3 I. 3 Because KH, HG, GB, BC are all equal, .·. ∆s ΑΚΗ, AHG, AGB, ABC are all equal. ... whatever multiple the base KC is of the base BC, the same multiple is AKC of △ ABC. Const. Similarly, whatever multiple the base CM is of the base base CM, ACM. AKC will be equal to, greater, or less than Now since there are four magnitudes BC, CD, △ ABC, and of BC and A ABC (the first and third) any equimultiples whatever have been taken, namely, KC and A AKC, and of CD and ACD (the second and fourth) any equimultiples whatever have been taken, namely, CM and ACM; and since it has been shown that if KC be equal to, greater, or less than CM, AKC is equal to, greater, or less than .. BC: CD = ABC: ACD. .. Again, because BC: CD= ACM; ABC: ACD; V. Def. 5 BC: CD = 2∆ABC: 2△ ACD V.15,11 = || EC : || CF. I. 41 COR. 1.-Triangles and parallelograms that have equal altitudes are to one another as their bases. COR. 2.-Triangles and parallelograms that have equal bases are to one another as their altitudes. For each triangle or || may be converted into an equivalent right-angled triangle or rectangle with base and altitude = its base and altitude; and in these latter figures the bases and altitudes may be interchanged. 1. If two triangles or ||ms have the same ratio as their bases, they must have equal altitudes; if they have the same ratio as their altitudes, they must have equal bases. 2. The rectangle contained by two straight lines is a mean proportional between their squares. 3. A, B, and Care three straight lines; prove that A has to B the same ratio as the rectangle contained by A and Chas to the rectangle contained by B and C. 4. A quadrilateral is such that the perpendiculars on a diagonal from the opposite vertices are equal. Show that the quadrilateral can be divided into four equal triangles by straight lines drawn from the middle point of the diagonal. 5. AB is || CD, and AD, BC are joined, intersecting at E; prove AE: ED = BE : EC. 6. Triangles ABC, DEF have A = 1 D, and AB = DE; prove △ABC: DEF = AC: DF. 7. AD, BE, CF drawn from the vertices of ABC to the opposite sides are concurrent at 0; prove BD : DC = ∆ AOB : △ AOC, CE : EA = △ BOC : △ BOA, AF : FB = ∆ COA : △ COB. 8. E is the middle point of AD, a median of △ABC; BE is joined and produced to meet AC at F. Prove CF = 2 AF. 9. ABC is any triangle; from BC and CA are cut off BD = onefourth of BC, and CE = one-fourth of CA. If AD, BE intersect at O, prove that CO produced will divide AB into two segments in the ratio of 9 to 1. 10. Perpendiculars are drawn from any point within an equilateral triangle to the three sides. Prove that their sum is constant. 11. Triangles and ||ms are to one another in the ratio compounded of the ratios of their bases and altitudes. PROPOSITION 2. THEOREMS. If a straight line be drawn parallel to one side of a triangle, it shall cut the other sides, or those sides produced * proportionally. Conversely: If the sides or the sides produced be cut proportionally, the straight line joining the points of section shall be parallel to the remaining side of the triangle. AAX C (1) Let DE be drawn || BC, one of the sides of ABC: it is required to prove that BD : DA Join BE, CD. Then A BDE = = CE: EA. CDE, being on the same base DE, and between the same parallels DE, BC; I. 37 (2) Let BD: DA = CE : EA, and DE be joined : it is required to prove DE || BC. * This useful extension was introduced by Robert Simson. Now these triangles are on the same base DE and on the same side of it; ... DE is || BC. I. 39 1. The straight line which joins the middle points of two sides of a triangle is || the third side. 2. The straight line drawn through the middle point of one of the sides of a triangle and || another side will bisect the third side. 3. Any two straight lines cut by three parallel straight lines are cut proportionally. (Euclid, Data, Prop. 38.) 4. Any straight line drawn || the parallel sides of a trapezium divides the non-parallel sides, or those sides produced proportionally. 5. In the figures to the proposition, if DE be || BC, prove BA : AD = CA: AE, and conversely. 6. ABC is any angle, and P a given point within it; draw through Pa straight line terminated by BA, BC, and bisected at P. 7. In the base BC of △ ABC any point D is taken, and DE, DF, drawn || AB, AC respectively, meet the other sides at E, F: prove △ AFE a mean proportional between As FBD, EBC. Examine the case when D is taken in BC produced. 8. ABC, DBC are two triangles either on the same side, or on opposite sides of a common base BC; from any point E in BC there are drawn EF, EG respectively || BA, BD, and meeting the other sides in F, G. Prove FG || AD. Examine the case when E is taken in BC produced. 9. ABC is any triangle; D and E are points on AB and AC such that DE is || BC; BE and CD intersect at F. Prove that △ ADF = ∆ AEF, and that AF produced bisects BC. Examine also the cases when D and E are on AB and AC produced. 10. Prove the following construction for trisecting a straight line AB in G and H: On AB as diagonal construct a ||m ACBD; bisect AC, BD in E and F. Join DE, FC cutting AB in G and H. 11. AB is a straight line, and C is any point in it; find in AB produced a point D such that AD: DB = AC : CB. |