Therefore BA: AC=BD: DC. = Cor. If AD, AE bisect the interior and exterior angles, at A, BA:AC= BD: DC, : also BA: AC= BE: EC. BE: EC= BE - ED: ED - EC. or Harmonic Section. THEOREM XXI. If a given straight line BC be divided in any ratio not equal to unity in the point D, another point E may be found in BC produced such that BE: EC= BD: DC. H B Draw any line BF. Join FD and produce it, through C draw G CII parallel to BF, meeting FD in G. Make CH = CG. Join FH. Then since BF is not equal to CII, BC, FI being produced must meet in some point E. Then BE: EC = FB:HC (IV. 2. = FB: GC =BD:DC. (iv. 2. Tie line BC is said to be harmonically divided. THEOREX INII. I be the middle point of a line AB which is divided Harmonicair in Canin, then saatke Spare on AN be qui tende maande V. 17. = THEOREMS. 1. AB, AC, AD, are lines drawn through A, EFG, KHL are parallel lines meeting them. Shew that EF: FG=KH : HL, and find all the proportions which the lines of the figure afford. 2. Any three lines are cut by three parallel lines, shew that they are divided proportionally. 3. AB, AC are drawn through A ; from B draw BC to any point C on AC, and from C draw CD to any point Don AB. Draw DE parallel to BC, and EF parallel to CD. Shew that AD is a mean proportional between AB and AF. 4. The distance of à point P from a given line AB is always in a constant ratio to its distance from another line AC; find the locus of P. 5. From points on the side of an equilateral triangle at distances 2, 4, 8 from one of the base angles perpendiculars are let fall on the base. Find the lengths which they intercept. 6. ABC, ABC" are triangles having equal angles at B, and at the angles C and C supplementary. Then BA: AC= BA: A'C'. B C" G Place the triangles so that their equal angles coincide, and through A draw AG parallel to A'C'. Then AGC= A'C'C= ACG, and AG = AC. But (IV. 2. Therefore BA: AC=B': 'C". i If two triangles have one angle equal, and the sides about a second angle proportional, their remaining angles are etter equal or supplementary. Nate the same construction as in 6 Sew that AC=AG. Tren AC eher aretes with AG, or is equally remote from the perpendicar oa the cate sie. Compare I Am. is case, and Cerers Ii,2. Een 40363 be ea both acute or both ob138, or one of them be a sale, the triangles are Santhat!!=139 (21. 10, IV. 3. In: Constancion s show that the singles ABD, 4422, C2 portiones also that 13:29=30, 01. 23:50 OD C. 11. The somierdie arse ha mor the remenicular of nr. pomilate sancis is to that of the side as 3 10 4. ihr entre Race is radius. On AO diameter anane cizale domhed, and any common charadas r. shant Show at the garments which is cus of trom zhe me erzins gate in the ratio of 4 10 1. Ti mer le numed chai segmens thich contain equal angies ar similar figurs 13. AB is the diameter of a circle, CD a perpendicular upon it from any point in the circumference. The semicircle on AB is equal to the semicircles on AC and CB together with the circle on CD. = 14. ABC is a triangle, D, E, F, the middle points of BC, CA, AB. 1. If AD = BE then A= B. middle point is AD. also AD. 4. The triangle constructed with the sides AD, BE, CF is to the original triangle as 3 to 4. 5. The sum of the squares on AD, BE, CF is to the sum of the squares on the sides of the triangle as 3 to 4. 15. A triangle may be divided into three equal parts by lines drawn from a point within it to the angles. 16. ABC is an equilateral triangle, AD the perpendicular from A, DG the perpendicular from D on AB. Shew that GB is one-fourth of AB. Determine the ratio of the squares on AG and AD. 17. ABC is a triangle, A a right angle, AD the perpendicular from A on BC. Shew that BC2 : BA?: AC as BC: BD: DC. 18. ABC is any triangle, AD the perpendicular on BC; shew that AB? - ACP = DB - DC. 19. Find the area of the triangle whose sides are 17, 15, and 8. |