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whence the rectangle BG, BE is equal to the rectangle BF, BD; a circle may therefore be described through D, E, G, F; whence DEG, DFG are equal to two right angles, i. e. to DEG, DEB; and .. DFG is equal to DEB, or to DAB and ABG; and .. the arc AB equal to the arc BC, which is absurd, unless the triangle be isosceles. Hence .. BG does not bisect the angle; and no other but BD can bisect it.

(34.) In any triangle, if perpendiculars be drawn from the angles to the opposite sides; they will all meet in a point.

Let ABC be any triangle; and AF, CD perpendiculars drawn upon the opposite sides, intersecting each other in G. Through G draw BGE; it is perpendicular to AC.

E

Join FD; and about the trapezium BFGD describe a circle. The triangles ADG, GFC being equiangular,

AG GC GD: FG,

whence also the triangles AGC, FGD are equiangular; and .. the angle ACD=DFG=ABE; and the angle BAC is common to the two triangles ABE, ACD; .. the angle AEB=ADC, i. e. it is a right angle, and BE is perpendicular to AC.

(35.) If from the extremities of the base of any triangle, two perpendiculars be let fall on the line bisecting the vertical angle; and through the points where they

meet that line, and the point in the base whereon the perpendicular from the vertical angle falls, a circle be described; that circle will bisect the base of the triangle.

H

Let ABC be a triangle, whose vertical angle B is bisected by the line BD, on which let fall the perpendiculars AF, CE. From B let fall BG perpendicular to AC; a circle described passing through E, F, G will also bisect AC.

P

G

About the triangle ABC describe the circle ADB; and from D draw a diameter which will bisect AC in H. Now since the angle AID is common to the triangles AIF, HID, and the angles AFI, IHD are right angles, the triangles AIF, HID are similar. In the same manner BIG, CEI are similar. Whence

HI: ID :: IF: IA,

and IG IB :: IE: IC,

:. HI× IG : ID × IB :: IF × IE: IA× IC, and since ID × IB=IA × IC, .. HI× IG= IF × IE, or a circle passing through E, F, G will pass through H (vi. 13.), and .. bisect the base AC.

(36.) If from one of the angles of a triangle a straight line be drawn through the centre of its inscribed circle, and a perpendicular be drawn to this line from one of the other angles; the point of intersection of the perpendicular, and the two points of contact of the inscribed circle, which are adjacent to the remaining angle, are in the same straight line.

Let ABC be a triangle, and O the centre of its inscribed circle. From B draw BOD through the centre;

and from Clet fall CH perpendicular to it. Let the circle touch the sides. of the triangle in Fand G; join HG, GF. HGF is a straight line.

Join OG, OC; and let a circle be described about the triangle ABC; join CD. The triangles OGE, CHE,

[blocks in formation]

having the vertical angles at E equal, and OGE, CHE right angles, are similar,

.. CE: HE OE: EG,

and the rectangle CE, EG is equal to the rectangle OE, HE; whence a circle will a circle will pass through the points C, O, G, H, the angle COH = CGH. Again (vii. 20.) CD= DỌ, and AG=AF, also the angle CDO=GAF, .. COE=AGF; whence CGH=AGF; and CG, GA are in the same straight line, .. FG, GH are in the same straight line.

(37.) If from the three angles of any triangle three straight lines be drawn to the points where the inscribed circle touches the sides; these lines shall intersect each other in the same point.

a

Let ABC be a triangle, in which circle is inscribed, touching the sides in E, F, D. Join AF, CE, cutting each other in O. Join BO, and produce it; it will pass through D.

For if not let it pass through some other point K; draw EG, EH respectively parallel to AC, BC. Then the triangles OEH, OFC being similar,

KK

[blocks in formation]

OE EH :: OC: (CF) CD;

in the same manner,

EG EO: KC CO,

.. EG: EH :: KC: CD.

Again, since EH is parallel to BF,
AE

EH: AB : (BF=) BE :: AK: EG

.. AK: AE :: EG: EH :: KC : CD,
or AK: AD :: KC : CD;

.. AK+ KC: AD+ DC :: AK: AD;
whence AK=AD, and K coincides with D.

(38.) If three circles touch each other, two of which are equal; the vertical angle of the triangle formed by joining the points of contact, is equal to either of the angles at the base of the triangle, which is formed by joining the centres.

Let three circles, whose centres are A, B, C, touch each other in the points D, E, F; and let the two circles, whose centres are A and B, be equal. Join AB, BC, CA, ED, DF, FE; the angle EDF is equal to either of the angles at A or B.

E

Since AE is equal to BF, the sides of the triangle ACB are cut proportionally, .. EF is and the angle FED is equal to EDA.

parallel to AB,

Now since CA

is equal to CB, the angle at A is equal to the angle at B ; but AD, AE are each equal to BF, BD, .. DE is equal to DF, and the angle DFE FED= EDA=AED ;

=

whence the angle EDF=DAE=DBF.

(39.) If three equal circles touch each other; to compare the area of the triangle formed by joining their centres with the area of the triangle formed by joining the points of contact.

Let three equal circles, whose centres are A, B, C, touch each other in D, F, E. Join AB, BC, CA, ED, DF, FE.

Since the circles are equal, their radii are equal, .. the sides of the triangle

E

F

B

ACB` are cut proportionally, and DF is parallel to AC, and DE to BC; .. AEFD is a parallelogram, and the triangle DEF is equal to ADE. In the same manner it may be proved to be equal to each of the triangles DBF, FCE; and .. it is equal to one fourth of ABC.

(40.) If four straight lines intersect each other, and form four triangles; the circles which circumscribe them will pass through one and the same point.

Let the lines AB, AC, DE, DC form the four triangles ABC, AEF, DCE, DBF; and let the circles circumscribing AEF, DBF, intersect each other in G; the circles circumscribing the triangles ABC, DEC will also pass through G.

Join GA, GE, GF, GB, GD.

Because the points

G, F, B, D are in the circumference of a circle, the angle GDB=GFA = GEA, i. e. GDC GEA, and =

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