Suppose ACD to be the required triangle, and IB to be the difference between the segment DI and its conterminous side DC, and IF the difference between the seg. AI and its conterminous side AC. Because AC is to AF, the angles ACF and AFCare; for the same reason the angles DCB and DBC are =.

=

Therefore it is evident, if you erect a perpendicular IC to the given one, from any point in an indefinite" right line, and inake IF,IB to the given differences, join BC and FC, make the angle BCD to CBD and the angle FCA to CFA, that ACD is the required triangle.

PROP. 55. THEOR.

No two lines from any points (D,E) in the sides of a triangle, to the opposite angle, can mutually bisect one another.

Fig. 46.

If one of them BE is bisected in F, the other CD is not; for if it was, then, in the triangles DFB,CFE the sides DF and FB would be respectively to CF and FE, and the angle CFE being to the angle DFB, the angle FDB would be to FCE; but those angles are alternate, .. AB would be parallel to AC, which is absurd.

PROP. 56. THEOR.

If any point (H) bé taken, without a parallelogram (AF), and lines be drawn from this point to each extremity of the sides of the parallelogram, the difference between the triangle (HAB) formed by one pair of lines (HA,HB) and the triangle (HDF) formed by the other pair of lines (HD,HF) is equal to half the given parallelogram.

Fig. 39.

Draw through the given point LK parallel to AB and produce DA and FB to meet it.

Then the triangle HAB is half the parallelogram KB, and the triangle HDF is half the parallelogram KF: but the difference between the parallelograms KB and KF is AF, ... the difference between their halves or between HAB and HDF is half AF.

PROP. 57, PROB.

Two right lines (AB,CB) being given in position, it is required to place a line between them equal to a given right line, and forming with them an isosceles triangle.

Fig. 47.

Draw through the point B, any line, EF, making angles with AB and CB, make BE and BF each to half the given line, through E draw EA (at right angles to BE) to meet BA, and through F draw FC parallel to EA; the line joining AC is the line required to be

`drawn.

For, in the triangles BAE,BCF, the angles AEB, ABE of the one are respectively to CFB,CBF of the other, and the side BE to BF, ... BA is to BC (by 26. 1.) .. the triangle BAC is isosceles; and AC is. to EF (by 33. 1.) and .. to the given line.

PROP. 58, THEOR.

If two friangles (ABC) and (DEF) be so situated, that the sides (AB and DB) passing through one extremity of the common base, being produced, make the angle (F) equal to I), and the side (DC) being produced through the other extremity, makes the angle (L) equal to (E), the angle (X) shall be double of the angle (Y).

Fig. 48.

For L is to Y and I, .. E is to Y and I, ... Y is the difference between 1 and E, also O is to Y and E, .. twice Y is the difference between O and I, but I is to K, and O is to X and K, ... X is double of Y、

PROP. 59, THEOR.

In any triangle, (ABD,) if the angles (ABD,ADC) be bisected, and the lines bisecting them be produced till they meet (in G,) the angle [BGD] is half (BAD.)

Fig. 41.

For the whole angle ADC is (prop 16. 1. Elr.) .. half of ADC, to half ABD, (viz. GBD,) and half = to DBG and BGD, ... BGD is

to ABD and BAD, (viz. GDC,) is = BAD, but GDC is to half BAD.

PROP. 60, THEOR.

If an angle (ABC) at the base of an isosceles triangle be bisected, and the right line (CD) bisecting it be produced to meet a right line (AD) parallel to the base, it makes with it an angle equal to half of one of the base angles,

Fig. 49

Produce the side BA through the vertex.

Then the angle EAC is evidently bisected by AD, .'. (by prop. 59) the angle ADB is half of the angle ACB, but half ACB is to ABD, .'. &c. &c.

=

PROP. 61. PROB.

In proposition fifty-nine of those deducibles, in what case will the right line (AD) connecting the vertices be parallel to the base.

Suppose AD is parallel to BG, then the angle ADB is to the angle DBC and .. to DBA,.. AD is - to AB: also the angle ADC is = to DCG and .. to ACD, .. AD is to AC,.. the triangle ABC will in this case be isosceles.

PROP. 62. THEOR.

If a right line (AD) be drawn bisecting an angle (ABC) at the base of an isosceles triangle, to meet a line parallel to the base, the right line [DC] drawn from its extremity to the external angle (ACG) bisects it.

Fig. 49.

For the angle ADB is to DBC and .. to ABD, .. AD is to AB, .. AC is to AD, and the angle ADC to the angle ACD, but ADC is to DCG,.. DCG is = to DCA, .'. &c.

Cor. It immediately appears from this, that in the case of an isosceles triangle; AD is par. to BG, if the angles are bisected by BD and CD. For if not draw AI par. to it, join IC. IC bisects ACG, but DC also bisects it, which is absurd.

PROP. 63, PROB.

In proposition fifty-nine, when will the figure [ABCD] be a parallelogram.

Because ABCD is a parallelogram, AD is to BC, (by prop. 34. 1. Elr.) it is also to AC and to AB, (by the proof of foregoing prop.) .. AB, BC and CA are to one another, .. the triangle ABC is equilateral. If the triangle ABC be equilateral ABCD is a parallelogram.

Since the angle ACG is bisected, DCG is an angle of an equilateral triangle, but DCG is to DBC,BDC; and DBC is an angle of an equilat. .. CDB is an angle of an equilat... BC is to CD: and in the triangles CBD, ABD the sides CD, BD are respectively = to AB,BD, and the angle CBD = to ABD, ... CD is

to DA, .'. BC, CD, DA and AB are = to one another,.. the figure is a parallelogram,

PROP. 64, PROB.

Given the base of any triangle, the area and the line bisecting the base to construct it.

Divide the given area by half the base and you have the altitude; erect from the middle point of the base a perpendicular to the altitude, and draw through its extremity a line parallel to the base; inflect from the middle point of the base to this parallel, a line to the given line bisecting the base, and connect its extremity with the extremities of the base, thus the required triangle is formed; evident from prop. 41. 1. Elr.

PROP. 65. THEOR.

In a right angled triangle, [CBA,] if one angle at the hypothenuse be double of the other, the hypothenuse is double of the side subtending the lesser angle.

Fig. 8.

Bisect CA in D, join DB.

Because BCA is double of BAC, BCA is an angle of 60 degrees, and because CBA is a right angled triangle, DB is to DC (by prop. 8 dedu.).. DBC is an angle of 60 and .. CDB .. the triangle ACD is equilateral, .. BC is to CD and.. of CA.

PROP. 66, THEOR.

If in an equilateral triangle, [BAC,] if [BD] is perpendicular to AC, and DO to CB, and IO equal to IB, the triangle [DOI] is equilateral.

Fig. 21.

Because DOB is a right angled triangle, having the angle OBD an angle of 30 degrees, the angle BDO must be an angle of 60, .. it is evident (from prop. 60 of dedn.) that if IO is to IB, the triangle IQD is equilateral.