same manner CF-FB,.. the radii of the circles are all 1 equal. And the circles touch each other in D, E, F, because the lines joining the centres pass through those points. (51.) Every thing remaining as in the last proposition; to describe a circle which shall touch the three circles. Bisect the angles CAB, CBA (see last Fig.) by the lines AO, BO meeting in 0. O is the centre of the circle required. H :. Join OC. Since the angle CAB=CBA, :. their halves are equal, or OAB=OBA, .. OA=OB. Also since CA= AB, and AO is common, and the angle CAO BAO, :. CO=OB; and the three lines OA, OB, OC are equal; parts of which GA, HB, CI are equal; .. OG, OH, OI are equal; and a circle described from O as a centre, with a radius equal to any one of those lines, will pass through the extremities of the other two, and touch the circles in the points G, H, I; because the lines joining the centres pass through those points. In nearly the same manner a circle may be described which shall touch the three circles on the opposite circumferences. (52.) To determine how many equal circles may be placed round another circle of the same diameter, touching each other and the interior circle. Let A be the centre of the interior circle, and AD its radius. Describe (vi. 50.) the circles DF, EF touching the circle DE, and each other. Then the angle at A being one third part of two, or one sixth part of four right angles, subtends an arc ED equal to one sixth of the whole circumference. And the same being true of every other contiguous circle, the number of circles which can be described touching each other and the interior one will be six. (53.) From a given rectangular parallelogram to cut off a gnomon, whose breadth shall be every where the same, and whose area shall be to that of the parallelogram in any given ratio. Let AC be the given parallelogram. Produce AB to D making BD = BC. On AD describe a semicircle, and produce CB to E; and let the ratio of the part to be cut off, to the whole, be that of 1: n. Make BE : BF n n-1; and between BE and BF. H F E 0 B take BG a mean proportional Bisect AD in 0; and with the centre O, and radius OG, describe a semicircle HGI; AH= ID, will be the breadth of the gnomon. Make BL BI, and draw HK, LK parallel to the sides of the parallelogram; then AC HL in the ratio compounded of the ratios of AB : BH and BD : BI, : i. e. in the duplicate ratio of BE BE BF, i. e. in the ratio of n BG, or the ratio of n-1, .. the gnomon AKC is to AC in the ratio of 1: n. (54.) To describe a triangle equal to a given rectilinear figure, having its vertex in a given point in a side of the figure, and its base in the base (produced if necessary) of the figure. Let ABCDEF be the given rectilineal figure, and P a given point in CD, which is to be the vertex of the triangle, the base being in AF. Join CA, and draw BG parallel to it; join CG, PG, PF, PE. Draw CH parallel to PG. Join PH. Draw DI parallel to PE, meeting FE produced in I. Join PI; and draw IK parallel to PF, meeting AF in K. Join PK; HPK will be equal to ABCDEF. Since BG is parallel to CA, the triangles BAG, BCG are equal; the figure therefore is equal to GCDEF. And since GP is parallel to CH, the triangles GCP, GHP are equal. Again since DI is parallel to PE, the triangles PIE, PDE are equal; ... PDEF is equal to the triangle PIF, i. e. to the triangle PKF, since IK is parallel to PF; whence the whole figure EE ABCDEF is equal to the triangles PHG, PGF, PKF, i. e. to the triangle PHK. (55.) On the base of a given triangle, to describe a quadrilateral figure equal to the triangle, and having two of its sides parallel, one of them being the base of the triangle; and one of its angles being an angle at the base, and the other equal to a given angle. E H E G F I D Let ABC be the given triangle, AC its base. At the point C make the angle ACD equal to the given angle; and let CD meet BD drawn parallel to AC, in the point D. On BD describe a semicircle BED; draw AF parallel to CD, and FE perpendicular to BD; and with the centre D, and radius DE, describe the arc EG. Draw GH parallel to AF, and HI to AC; AHIC will be the figure required. Join HC. Since DG=DE, BD: DG :: DG : DF, .. (Eucl. v. 19.) BG: GF :: DG : DF :: HI : AC. Now BG GF :: BH: HA, : .. BH : HA :: HI : AC. But the triangles HCI, AHC are in the proportion of HI: AC, and the triangles BHC, AHC in the proportion of BH: HA, :. HCI : АНС :: ВНС : АНС, or HCI=BHC; .. ACH, and HCI together are equal to ACH, and BCH together, or AHIC=ABC. (56.) A trapezium being given, two of whose sides are parallel; to describe on one of those sides another trapezium, having its opposite side also parallel to this, and one of the angles at the base the same as the former, and the other equal to a given angle. Let ABCD be the given trapezium whose sides AD, BC are parallel. Join BD; and draw CE parallel to produced in E. it, meeting AB Then the trian E F B gles BCD, BED are equal; and .. the triangle AED is equal to ABCD. Hence (vi. 55.) a figure ADGF may be described equal to ADE, and .. to ADCB. (57.) If with any point in the circumference of a circle as centre, and distance from its centre as radius, a circular arc be described; and any two chords be drawn, one from the centre of the circular arc, and the other through the point where this cuts the arc, and parallel to the line joining the centres; the segments of each chord intercepted between the circumferences which are concave to each other, will be equal respectively to those of the other between the other circumferences. I B E D G C With any point C in the circumference of the circle ABC as centre, and radius CE equal to the distance from the centre E, let a circle DFE be described. Join CE, and draw any chord CFA; and through F draw HFG parallel to CE; then will CF Produce CE to B, = FH, and GF=FA. and join HE. And since HG is |