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drawn from these points to the radius ; the mirtilinear space cut off, shall be equal to the sector which stands on the arc between them.
Let two points C, E be taken at equal distances from A and B, the extremities of the quadrant AB; and let fall the perpendiculars CD, EF on A0. Join CO, E0; the figure CDFE is equal to the sector COE.
Since the arc AC=EB, the angle AOC=EOB= OEF, and the angles at D and F are right angles, and CO=OE, .. the triangle COD=EOF; from each of these take away OFH, ... DFHC=OHE; to each of these add CHE, and CDFE=COE.
(33.) If the arc of a semicircle be trisected, and from the points of section lines be drawn to either extremity of the diameter ; the difference of the two segments thus made, will be equal to the sector which stands on either of the arcs.
Let the arc of the semicircle ACB be divided into three equal parts in the points C, and D. From A the extremity of the diameter draw AC, AD; take ( the centre, and join OC, OD; the difference of the segments AC and ACD is equal to the sector COD.
Since the-angles CAD, DAB stand on equal circumferences, they are equal; but the angle DAO=ODA, .:: CAE=EDO; and CEA= OED,
OED, :. the triangles CAE, EOD are equiangular ; and since OE is drawn bisecting the vertical angle of the isosceles triangle
AOD, it bisects the base, .. AE=ED; and consequently the triangle AEC is equal to the triangle DOE; add to each CED, and CAD=COD.
(34.) If a straight line be placed in a circle, and on the radius passing through one extremity, as a diameter, another circle be described ; the segments of the two circles cut off by the above straight line will be similar, and in the ratio of four to one.
Let EC be a straight line placed in the circle ABC. Take O the centre, and join OC; and upon it describe a semicircle ODC. The segments EAC, DGC are similar, and in the ratio of 4 : 1.
Join OD, and produce it both ways to the circumference. Take F the centre of the semicircle ODC. Join OE, FD. Then ODC being a right angle, ED = DC, and OF=FC, :. (Eucl. vi. 2.) DF is parallel to EO;
and CE : EO :: CD : DF, and the segments EAC, DGC are similar;
whence EAC : DGC :: EC : CD :: 4 : 1.
Cor. The segment ADC is bisected by the circumference DGC.
(35.) If on any two segments of the diameter of a semicircle semicircles be described; the area included between the three circumferences will be equal to the area of a circle, whose diameter is the mean proportional between the segments.
On AD, and DC, segments of AC the diameter of the semicircle ABC let semicircles · AED, CFD be described; from D draw DB perpendicular to AB, and .. a mean proportional between AD and CD; the figure AEDFCB is equal to the circle described upon DB.
Join AB, BC. Since ADB is a right angle, the semicircle on AB is equal to those on AD and DB together; and the semicircle on BC is equal to those on BD and DC;... the semicircles on AB and BC, or the semicircle ABC which is equal to them, will be equal to the semicircles AED, DFC and the circle described upon DB. From these equals take away the semicircles AED, DFC, and the figure AEDFCB is equal to the circle described upon DB.
(36.) If the diameter of a semicircle be divided into any number of parts, and on them semicircles be described; their circumferences will together be equal to the circumference of the given semicircle.
Let AB the diameter of the semicircle ACB be divided into any numher of parts in the points D, E; and on AD, DE, EB, let semicircles be described; their circumferences are together equal to ACB.
For since the circumferences of circles are as their diameters,
ACB : AFD :: AB : AD
ACB : EHB :: AB : EB, whence ACB : AFD + DGE + EHB :: AB : AD+
[DE+EB, in which proportion the third term being equal to the fourth,
ACB= AFD+DGE+ EHB.
(37.) If two equal circles cut each other, and from either point of section a line be drawn meeting the two circumferences ; the area cut off by the part of this line between the two circumferences will be equal to the area of the triangle contained by that part and lines drawn to its extremities from the other point of section.
Let the two equal circles ADB, ACB cut each other in A and B; and from A draw any line AC, cutting the circles in D and C; join DB, BC; the figure Db BcCD is equal to the triangle DBC.
Take any points E, F in the circumferences AEB, AFB; join AE, EB, AF, FB. Since the arcs ADB, AFB are equal, the angles ADB, AFB are equal. But the angles AFB, AEB are equal to two right angles, and .:: to ADB, BDC; whence the angle BDC=AEB= ACB, and BD=BC; .. the segment Db B is equal to the segment BcC; to each of these add DbBC, and the triangle DBC is equal to Db BcCD.
Cor. If AE is a tangent to ADB at A; the area ADbBcCE A will be equal to the triangle ABE.
(38.) If two equal circles touch each other externally, and through the point of contact another be described with the same radius ; the area contained by the convex circumferences cut off from the touching circles, and the part of the third without them, is equal to the area of the quadrilateral figure formed by lines drawn from the points of intersection to the point of contact, and to the point where the third circle is cut by a tangent drawn to the point of contact of the two circles.
Let two equal circles touch each other in A; and through the point of contact let an equal circle ABC be described, cutting the former in B and C. Join AB, AC; and to the point A let a tangent AD be drawn ; join BD, DC. The area contained by AEB, AFC and the intercepted arc BDC is equal to the quadrilateral figure ABDC.
Since DA touches the circle AEB, the angle DAB is equal to the angle in the alternate segment, and .. equal to the angle in the segment BCA, i.e. equal to the angle BDA, whence BA=BD; .. the segment BEA is equal to the segment BGD; and AEBGDA is equal to the triangle ABD. In the same manner it may
be shewn that AFCHDA is equal to the triangle ACD; :. AEBGHCFA is equal to the quadrilateral figure ABDC.
(39.) If a straight line be divided into
two parts, and upon the whole and the two parts semicircles be described ; and from the point of section a perpendicular