Fig. 73.

In the triangles d CF, CA c the angles AC c, C F d are evidently to one another; and the angle FdA is

to the angles dC F, CFd and also to the angles d C'i, Cid,.. the angle Cid is to dC F, .. Cid is a right angle, eik is a right angle, for the same reason the angles iky, kge, gei are right angles, the figure eg ki is rectangular. Then in the triangles FC i, DFe the sides De, Fi are evidently, and since FC is bisected in & and u e parallel to Ci, Fi is bisected in e, for the same reason De is bisected in g,.. ie is to eg and to the opposite sides ik, kg,.. ikge is a

square.

And each of the triangles ae F, bg D, çk A, diC is the fourth part of each of the triangles FCi, CAk, A Dg, D Fe, .. the four small ones are and are together to the fourth part of the figure outside eikg: bisect ei in o then the triangle e o g is evidently the fourth part of eik g, and the triangles geo, Fea are evidently, to one another, .. the four Fae, C di, Ack, Dbg are together to eik g .. it is the fourth part of the figure outside itself, and.. the fifth part of the ' FCAD.

PROP. 124, THEOR.

In the theorem of Pappus, (vide Elrington, page 158,) the right line [B n] drawn from one extremity of the side [BA] of the given triangle [ABC] equal and paral lel to the right line [CV] joining the vertex and point of concourse of produced sides, shall fall on the produced side [OVv.]

Fig. 69.

For if it does not, let it fall at X draw X V.
Because BX is

VX is parallel to FB,

and parallel to CV, the right line (prop. 33, 1 Elr.) but VO is also

parallel to it which is absurd, it is.. evident that the right

ne must fall on OV.

PROP. 125, THEOR.

In the theorem of Pappus, if the given triangle (ABC) be right angled, the parallelogram (n BA r) described on the hypothenuse shall be a square, if those (FA,LB) described on the sides. be squares, (which shews that the 47th is a particular case of it.)

Fig. 69.

For in the triangles BCA,CLV the sides BC and CA of the one, are respectively to CL,LV of the other, and the angle BCA is to CLV, .. BA is to CV and .. to Bn,.. the figure n B Ar is equilateral; and because the angle VCL is to CBA the angles LCV, BCT are to a right angle, .. LCV, LCB and BCT are together to two right angles, .. TC,CV are in directum,.. TC is parallel to Bn, .. the angles n BC, BCT are =, .. the angle n BA is a right angle, for the same reason the angle B A r is a right angle,.. the figure BArn is a square.

PROP. 126, THEOR.

Prove the three angles of any triangle (ABC) equal to two right angles without producing a side.

Fig. 1.

Draw through the vertex A, any right line. FG parallel to the base- BC.

Then the three angles BAF, BAC, CAG are = to two right angles (prop. 13. I cor. Elr,) and because FG is parallel to BC the angles FAB,ABC are, also the angles GAC,ACB, .. ABC,BCA and CAB are togetlier to two right angles.

PROP. 127. PROB.

In the ninth proposition of first book of Elrington, if the diagonals (AF and ĎE) be equal, what will the measure of the given angle (BAC) be.

L

Fig. 22, plate 1 Elr.

If the right line FA is to 'DE it is to FE.. the angles FAE,FEA are, but the angle AFE is an angle of 30,.. each of the angles FAE,FEA is an angle of 75, for the same reason each of the angles FDA,FAD is au angle of 75, .. the given angle BAC is an angle of 150 degrees,

PROP. 128. PROB.

To construct an isosceles triangle whose vertical angle shall be four times one of its base angles.

Suppose that MIL is the required triangle having the angle MIL four times the angle IML.

Bisect the angle MIL by IP and produce it to O, draw to meet it MO,LO making the angles IMO,ILO each = IML or ILM. The angle PIM is to twice PMI, i. e. to PMO, and it is also to IOM,IMO, .. IOM,ÍMÒ ILO,IOL are to one another, the angles MOL, OLM, LMO are to one another, ... OML is an equi

lateral triangle.

If therefore you construct an equilateral triangle OML, bisect the angles OML,OLM, the right lines MI,LI bisecting them form with ML the triangle IML, having its vertical angle MIL four times one of its base angles. The demonstration is evident.

PROP. 129, THEOR.

If two right angled triangles (ACG,FDE) be similar, and if the right line (CB) drawn from the right angle (ACG) of one, perpendicular tothe hypothenuse (GÃ), be equal to a side (FD) of the other, the hypothenuse (FE) of the latter shall be equal to a side of the former.

Fig. 55.

For since the triangles are similar, the angles CAG, DFE are, also the angles DEF,CGA; and since CB is a perpendicular, the angle BCG is to GAC, ..

BCG is to DFE, also the angles CBG,FDE are= being right angles, and the perdendicular CB is to the side FD,.. (prop. 26, 1 Elr.) the triangles CBG,FDE are in every way, .. the side CG shall be to the hypothenuse FE.

PROP. 130, PROB.

To draw from the vertex of a scelene triangle (BCD) to the base a right line which shall exceed the less side (BC) as much as it is exceeded by the greater (CD).

Suppose that CG is the line required to be drawn, and that it exceeds BC as much as it is exceeded by CD; make CF = to CG; CF exceeds CB by a part to FD; make FE to DF, then CE is = to CB, ... ED is the difference between CD and CB, EF is half the

difference.

If therefore you make CE to CB and bisect the difference ED in F, EF is evidently to the line required to be drawn; .. from C as centre, and FS as radius, describe a circle, the point G where it cuts the base is the point to which the line is to be drawn.

PROP. 131, THEOR.

The right line (DF) drawn from the vertex of a scelene triangle (ADB) to bisect the base, is greater than the right line (DC) bisecting the vertical angle and produced to the base.

Fig. 17.

Since the angle DBA is greater than DAB, and DC bisects the vertical angle, CA is greater than CB, (prop. 19, dedu... DF falls between DA and DC. Then the angle DCA is to the angles CBD,BDC, .. to the angles CBD,CDA; and the angle BFD is = to the angles FDA, FAD, but FAD is less than FBD and FĎA less than CDA, .. BFD is less than FCD, .'. DF is greater than DC.

PROP. 132, THEOR.

The two sides (AD,DB) of any triangle (ADB) are together greater than double the right line (DC) bisecting the vertical angle and drawn to the base.

Fig. 17.

If the given triangle is isosceles, the line bisecting the vertical angle also bisects the base; then it is evident from (prop. 91 of deduc.) the sum of the sides AD,DB is greater than double DC.

But if the triangle be scelene, then the segment CA of the base adjacent to the greater side is greater than the segment BB, (prop. 19, deduc.) bisect AB in F, join F,D. Then FD is greater than CD, (prop. 130, deduc.) but the sides AD,DB are together greater than double of FD, .. they are much greater than the double of CD.

PROP. 133, PROB.

Given of any triangle, one side, an angle at it, and the sum of the other two sides, to construct it.

Suppose that ABD is the required triangle: let AB be the given side, ABD the given angle, and BD = to the given sum; join D,A.

Then, since BD is to the sum of BC and CA, CD is to CA,.. the angles CAD,CDA are.

If therefore you draw BA to the given side, make the angle ABD to the given angle, make BD = to the given sum; join DA and make the angle DAC to CDA it is evident that ACB is the triangle required to be formed,

PROP. 134. THEOR,

=

To find a point in the base (AD) of a scelene triangle (ABD,) through which if a line be drawn at right angles to one of the sides, (BA,) the segment of the base between that side and the required point, shall be equal to the segment of the perpendicular between the point and the other side (BD) produced,