95. If several right lines meeting, or intersecting each other in a point P, are point P, are cut by two parallel lines AB, CD; the intercepted segments will be respectively proportional: AG: CO GN: OI :: NB: ID, &c. For the triangles APG, CPO; GPN, OPI; NPB, IPD are respectively equi-angular, and therefore similar; Hence (94, corol. 1), AG: CO :: GP : OP :: GN: OI :: NP: IP :: NB : ID, &c. Therefore (31) AG CO: GN: OI :: NB: ID, &c. Corol. Hence it is evident, if AG, GN, &c. and CO, OI, &c. are not in the same continued right lines, but respectively parallel as before, that CO, OI, ID, &c. will be in the same proportion as AG, GN, NB, &c. A C A 96. The line BG bisecting the vertical angle ABC of the triangle ABC, divides the base AC into two parts having the ratio of the sides AB, BC: AB: AG :: BC: GC. Draw CD parallel to BG meeting AB produced in D. D B G Then because CD is parallel to BG, the angles BCD, GBC are equal (40). And the external angle CBA (or double the angle GBC) of the triangle CBD, is equal to both the angles BCD, BDC (41). Therefore the angles BDC, BCD are equal, and conse quently BD is equal to BC (46, corol. 2). But the triangles ABG, ADC are similar; Hence AB AG :: BD (BC): GC (94, corol. 1). Corol. Hence, if a line bisects the vertical angle of a triangle, the rectangle of either side and the alternate segment of the base, is equal to the rectangle of the other side and the remaining segment: AB x GC = AG x BC. 97. In a circle, if two chords AB, CD intersect each other, and their extremities are joined, the triangles PĊA, PBD will be similar; and the rectangle of the segments PA X PB equal to the rectangle of the segments PC × PD. For the angles PBD, PCA, standing on the same arc DA, are equal to each other (70). And the angles PDB, PAC, standing on the arc CB, are also equal. B D P A And the equal angles at P being common to both triangles, those triangles are therefore equi-angular, and consequently similar; Hence PA PC:: PD: PB (94, corol. 1): Therefore PA x PB PC × PD (89). Corol. 1. If one chord DC bisects the other AB at right angles, then DC is the diameter of the circle (65), and AP or PB is a mean proportional between DP and PC. Corol. 2. And if AD, AC are joined, the angle CAD is a right one (72); therefore the perpendicular AP let fall from the right angle on the hypotenuse DC, is a mean proportional between the segments DP, CP. Corol. 3. Hence also, if two lines AB, CD, intersect each other in the point P, and PA × PB = PD x PC; then a circle will pass through the points D, B, C, A. And the triangles PDB, PAC, will be similar. 98. If two right lines PA, PC from the same point P, intersect a circle, and the chords BD, AC are drawn; then the triangles BPD, CPA will be similar; and the rectangle PA x PB is equal to the rectangle PC × PD. For the sum of the two opposite angles EDC, BAC is equal to two right angles (74). And the angles BDC, BDP are together equal to two right angles. BD Therefore the angle BDP is equal to the angle BAC. And for the like reason, the angle DBP is equal to the angle DCA. And the angle P being common to both triangles, those triangles must be equi-angular, and consequently similar: Hence PA PC :: PD: PB, Therefore PA x PB PC x PD (89). 99. If PB be a tangent to a circle, and PC a secant; then the rectangle PC × PD is equal to the square of the tangent PB. Draw BD, BC. Then the angle PBD is equal to the angle PCB (73). And the angle ABC equal to the angle BDC (73). B THEOREMS. Therefore the angles ABC, BDC, being equal, their sup plements or the angles CBP, BDP must be equal. Consequently the triangles PDB, PBC are equi-angular: Hence PC: PB :: PB : PD. Therefore PC × PD PB2. 100. If two triangles BPD, bPd are similar; the bases BD, bd, and perpendiculars PA, Pa, are proportional: BD: bd :: PA: Pa. Because the angles BAP, buP are right ones, the triangles BAP, baP are also similar; Hence PB Pb :: PA: Pa (91, corol. 1), Therefore BD: bd :: PA : Pa (by equality). 101. The surfaces or areas of similar triangles are in the duplicate ratio (or as the squares) of their homologous sides. Let the triangles BPD, lpd be similar; and BN, bn, the squares on the sides BD, bd: Then, triang. BPD: triang. bpd :: square BN : square bn. P P B d n Suppose the perpendiculars PA, pa, are let fall on BD, bd, respectively; and join DG, dg. Then because the triangles BPD, BGD are on the same base BD, we have (87, corol. 2). Triang. BPD triang. BGD :: PA : BG (BD). And, triang. bpd triang. bgd: pa: lg (ld). But PA: BD :: pa : bd (100): Therefore (31), triang. BPD triang. BGD :: or triang. BPD square BN triang. bpd triang. bgd: triang. bpd: square bn; because the two squares must evidently have the same ratio as their halves. 102. All similar right lined plane figures (ABDNG, abdng) are to one another in the duplicate ratio, or, as the squares of their homologous sides (AG, ag). Draw GB, GD, gl, gd. Then the figures being similar, the angle A is equal to the angle a; and the including sides AB, AG; ab, ag, are proportional (90); therefore the triangles ABG, abg are equi-angular and similar (94, corol. 1). And if the equal angles ABG, abg are taken from the equal angles ABD, abd, the remaining angles GBD, gbd, must be equal. Hence AB ab :: BG: bg; AB ab: BD: bd; b D Therefore (31) BG: bg :: BD bd: consequently (94, corol. 1) the triangles GBD, gbd, are similar. And in the same manner it may be proved that the triangles GDN, gdn, are similar. Hence (101), triang.GAB:triang.gab::GB2:gb2::GBD:gbd::GD2gd: GDN: gdn; or GAB: gab :: GBD : gbd :: GDN : gdn: : And (94, schol.) GAB: gab :: GAB + GBD+GDN gab + ghd +gdn. But the antecedents together is the figure ABDNG, and the consequents the figure abdng; Therefore AG2: ag' (GAB: gab) :: ABDNG: aldng. To illustrate this by an example in numbers, suppose AG 10 feet, |