triangles FKB, ABC are equiangular and equal. E In like manner it may be proved that the triangles ABC, LCI are equiangular and equal. (43.) If from the angular points of the squares described upon the sides of a right-angled triangle perpendiculars be let fall upon the hypothenuse produced; they will cut off equal segments; and the perpendiculars will together be equal to the hypothenuse. Let FM, IN be drawn from the angles F, I of the squares described upon BA, AC, perpendicular to BC the hypothe F G H nuse produced; MB will be M B O From A draw 40 perpendicular to BC. Since FBA is a right angle, the angles FBM and ABO together are equal to FBM and BFM, :. ABO is equal to BFM; and the angles at M and O are right angles, and AB= BF, .. BM=AO, and FM=BO. In the same manner it may be shewn that CN=AO, and IN=CO; .. MB =NC, and FM and IN together are equal to BO and CO together, i. e. to BC. COR. The triangles FBM, ICN are together equal to ABC. (44.) If on the two sides of a right-angled triangle squares be described, the lines joining the acute angles of the triangle and the opposite angles of the squares will cut off equal segments from the sides; and each of these equal segments will be a mean proportional between the remaining segments. P H On AB, AC the sides of the right-angled triangle BAC, let squares be described, and BI, CF joined; the segments AP, AQ are equal, and each of them is a mean proportional between BP and CQ. Since 4Q is parallel to HI, and AP to FG, BH : HI :: BA : AQ, B and (CA=) III: CG :: AP (FG=) AB, : and BH being equal to CG, AP=AQ. Again, the triangles BPF, ACP being similar, as also ABQ, ICQ, BP: (BF=) AB :: AP : AC, and BA : AQ :: (IC=) AC ; CQ, .. ex æquo BP : (AQ=) AP :: AP.: CQ. (45.) If squares be described on the hypothenuse and sides of a right-angled triangle, and the extremities of the sides of the former and the adjacent sides of the others be joined; the sum of the squares of the lines joining them will be equal to five times the square of the hypothenuse. Let squares be described on the three sides of the rightangled triangle ABC; join DF, EI; the squares of DF and EI together are equal to five times the square of BC. Draw FK, IL perpendicular to DB, EC produced, and AM to BC. The angle FBK is F K H B M C D equal to ABC, and the angle at K to the right angle AMB, and FB=BA, .. BK=BM. CL=CM. In the same way, Now (Eucl. ii. 12.) FD = DB2 + BF2+2 DB × BK =BC2+BA'+2 BC × BM, and EI-BC' + CA +2 BC × CM, ... FD2 + EI=2 BC2 + BA + AC2 + 2 BC × BM+ [2 BCX CM =2 BC+BC2+2 BC2=5 BC2. (46.) If a line be drawn parallel to the base of a triangle, and terminated in the sides; to draw a line cutting it, and terminated also by the sides, so that the rectangle contained by their segments may be equal. Let ED be parallel to CB the base of the triangle ABC; from D draw DF, making with AC (produced if necessary) the angle DFE equal to ABC, and draw any line GH parallel to FD, cutting ED in I; the rectangle EI, ID is equal to the rectangle GI, IH. For the angle AGH= AFD=ABC =ADE, and the vertical angles at I are equal, .. the triangles GEI, HID are equiangular; and HI ID :: IE: IG, ..the rectangle EI, ID is equal to the rectangle HI, IG. (47.) If the sides, or sides produced, of a triangle be cut by any line; the solids formed by the segments which have not a common extremity are equal. Let ABC be a triangle having the sides (produced if necessary) cut by the line DEF; then AF × CD × BE = AE × DB × CF. X F GE D B EG Draw BG parallel to AC; the triangles AEF, BEG will be similar, as also CDF, BDG; .. AF AE :: BG: BE, : :. AF× CD: AEx CF :: BD: BE, (48.) If through any point within a triangle, three lines be drawn parallel to the sides; the solids formed by the alternate segments of these lines are equal. Through any point D within the triangle ABC, let HG, EF, IK, be drawn parallel to the sides; then ID x DGx DF-ED× DK× DH. Since the lines are drawn parallel to the sides, the triangles IED, GDK, HDF are similar to ABC, and to one another; .. ID: DE: AC: CB GD: DK :: AB AC DF: DH :: BC: AB, whence ID x DG × DF: DE × DK × DH :: AC× ABX BC BC× AC × AB, i. e. in a ratio of equality. (49.) If through any point within a triangle lines be drawn from the angles to cut the opposite sides; the segments of any one side will be to each other in the ratio compounded of the ratios of the segments of the other sides. B H D Through any point D within the triangle ABC, let lines AE, BF, CG be drawn from the angles to the opposite sides; the segments of any one A of them as AC, will be in the ratio compounded of the ratios AG GB, and BE : EC. C Draw IBH parallel to AC, meeting AE and CG produced in H and I. Then the triangles GCA, GBI, and EAC, EBH, as also ADF, BDH, and FDC, IDB, are respectively equiangular, |