Therefore at one extremity K of the given sum make the angle CKB to half the vertical angle, and from the other extremity C inflect the right line CB = to the given base, make the angle KBA to AKB, it is evident that CAB is the required triangle.

PROP. 146. THEOR.

If the sides (EB,GA) of a square described on the hypothenuse of a right angled triangle, be produced to meet the sides (DF,K) or those produced if necessary, the triangles (B On, r DA) cut off by them are equal and equiangular to the given triangle!

Fig. 69.

Because the angle n B A is a right angle, and also the angle OBC, if.. the angle n BC is taken away the angles O Bn. CBA remain to one another, .. in the triangles O Bn, CBA the angles O Bn, BOn of the one are respectively to CBA, BCA of the other, and the side OB is to BC, the triangles O Bn, CBA are = and equiangular; in like manner the triangles DrA, CBA can be proved equiangular and

=

PROP. 147. THEOR.

The right line (BD) joining the acute angle (CBA) of a right angled triangle, with the opposite angle (FDA) of a square described upon the side subtending this acute angle, will cut from the triangle an obtuse angled triangle (Be A) which shall be equal to that (eCF) cut off from the square, by à line (e F) drawn from the intersection of the connecting line with the side to the angle (BFD) opposite said connecting

1

line.

Fig. 69.

Draw the diagonal CD.

Then since the triangles Ce D, C e F are on the same base Ce and between the same parallels they are =, and also because the triangles DAB,DAC are on the same

are,

base DA and between the same parallels AD,FB they if .. the common part Ae D is taken away the remaining triangles D Ce, A e B shall be,.. A e B is to e FC: and the angle A e B is evidently obtuse since it is greater than e C B which is a right angle.

PROP. 148, PROB.

Given the lengths of the three lines which drawn from the angles bisect the opposite sides to construct the triangle.

Fig. 57

Construct a triangle OML whose sides shall be double the given bisectors; bisect MO in D, join DL, make LI 3rds of LD, then I'D is LI, produce ID until DB to it, join IO and BO: BOI is the required tri

is

=

angle. For DO is evidently to one of the given bisectors, since it is MO, which is double one of them; draw I e parallel to LO, then in the triangle BLO the side BL is bisected in I and I e drawn par. to LO; the side OB is also bisected in e, and I e is to half LO, (prop. dedu.)... Ie is to an other of the bisectors, .. it is evident that the right line Bg drawn through their point of concourse bisects IO (prop. dedu.) and must be = to the third bisector.

PROP. 149. PROB.

Given any quadrilateral figuré (GKM L it is required to bisect it by a right line drawn from a given point (A) in one of its sides.

Fig. 51.

Connect the given point A with the opposite angle GKM; bisect the figure AKML by A e, (prop. dedu.) upon AE and at the side of it at which the greater part of the given figure lies, describe a triangle A & M = to AKG, bisect this triangle by the right line A o; this right line A o evidently bisects the given figure.

DBb, b BA, Ad C, d CD are together

to half the sum of the triangles HEF,IEF, .. HEF + IEF + twice ABDC is to twice Ab Dd or to EHIF, but IGH + HGE + EGF+ FGI is also to EHIF, IGH and EGF are to twice EGF + twice ABDC, .. HGI - EGF is to twice ABDC, .. ABDC is to half the difference of the triangles HGI,EGF.

And of first

End of First look