PROP. 67, PROB.

To determine a point within a triangle from which the perpendiculars let fall on the sides shall be equal.

Because DE is to DF and DB common, and the angle DEB = to DFB, and the anglés EDB,FDB of the same affection, the angle EBD is to FBD), for the same reason the angles FCD,GCD are = and also

GAD and EAD.

Therefore, if you bisect each of those angles, the point D, in which the lines bisecting them meet, is the required point. For in the triangles DBF,DBE the side DB is common, and the angles DFB,DBF of the one respectively to DEB,DBE of the other, .. DF is to DE: for the same reason DF is = to DG, ... &c. &c.

PROP. 68. THEOR.

If two sides (CD,EF) of a quadrilateral figure be раrallel, the triangle (ICE) contained by either of the other sides, and the straight lines drawn from its extremities to the point of bisection of the opposite side, is half the figure.

Fig. 51.

Draw GH through the point of bisection I parallel to CE and draw FH. Then the triangle GID is evidently

to HIF, the parallelogram CH is to the given figure ED, but the triangle CIE is to half the parallelogram and .. to half the figure.

PROP. 69. PROB.

To bisect a given quadrilateral figure (GE) by a right line drawn from one of its angles (C.)

Fig. 52.

Draw diagonals CD,GE and bisect the diagonal GE opposite the angle C in O, through O draw LI parallel

to CD, the right line drawn from C to I bisects the given figure, join CO and DO.

Because GO is to OF the triangles GOC and EOC are, for the same reason the triangles GOD and EOD are,.. the figures GCOD and CODE are =; and the triangles COD,CID are= .. if the common part OFI be taken away the remainders COF,FID are,.. the triangle CIE is to the figure CIDG.

PROP. 70. THEOR.

If two sides (CE and EB) of any parallelogram be bisected, (in G and 1,) and lines drawn from the points of bisection to the opposite angle, they trisect the diagonal (CB) subtending this angle.

Fig. 15.

For in the triangle DCE, since the sides DE and EC are bisected in A and G, the lines CD and AC drawn to the opposite angles cut one another in a point of trisection (prop. 24 of dedu.).. AP is the third part of AC, it is.. half of PC, for the same reason AO is half of OB; but AB is to AC, and PO is the third part of their sum, .. it is evident that OB and PC are each .. &c.

PROP. 71, THEOR.

to PO,

If in a triangle (ABC) a line (BD) drawn from the vertical angle to the point of bisection of the base, be greater than half the base, the vertical angle is acute; and if the line be less than half the base, it is obtuse, (as in the triangle AEC.)

Fig. 19.

Because DB is greater than DA the angle DBA is greater than DAB, for the same reason the angle DCB is greater than DBC, .. the two BAC,BCA are greater than ABC,.. ABC is less than a right angle. Thus it can be proved that AEC is greater than a right angle, since DE is less than DA.

PROP. 72, PROB.

Given in a right angled triangle, one side and the sum of the hypothenuse and the other side, to construct

Suppose ABC to be the required triangle: AB is = the given side, and AD the given sum; join DB. Since BAC is a right angle, BC is the hypothenuse, and it is

to CD, (the difference between the second side AC and the given sum,) .. the angles CBD and CDB

are.

Therefore it is evident, if you form with the given side BA and given sum AD, a right angle BAD, join DB and make the angle DBC to the angle CDB, that BAC is the required triangle.

PROP. 73, PROB.

Given in any triangle the base, the difference of the sides, and the difference of the base angles to con

struct it.

Fig. 3.

ANALYSIS.

Suppose ADB to be the required triangle. Let AB be to the given base, and BC the given difference. Join CA. Because CB is the difference between AD and DB, AD is: to DC,.. the angles ACD,CAD are =: also because CB is the difference, the angle CAB is to half the difference of the base angles.

Therefore, it is evident, if from one extremity (A) of the given base, you draw AC making the angle CAB= to half the difference of the base angles, from the other extremity B, inflect BC to the difference of the sides, produce BC, and through A draw AD, (to meet it,) making the angle CAD to ACD, that ADB is the required triangle.

PROP. 74, PROB.

To bisect a given parallelogram (EA) by a right line drawn from a given point (0) reithout it.

Fig. 4.

Bisect the diagonal AE and draw OL through the point of bisection, it bisects the parallelogram.

For the triangles DAB,BED are, also the triangles DGK,BGL are,.. the figure EKLB is figure LADK.

PROP. 75, PROB.

to the

To find two equal squares whose sum shall be equal to the sum of two unequal squares.

Fig. 54.

With the sides (GK,GL) of the unequal squares form a right angle KGL; draw the hypothenuse KL, and on it describe an isosceles right angled triangle KML; the sum of the squares of KM,LM is to the sum of the squares of KG,GL.

For the squares of KG,GL are to the squares of KL, (prop. 47. 1. Elr) also the squares KM,ML are = to the square of KL, .'. &c.'

PROP. 72, THEOR.

It is required to draw, from two given pointh A,B) to the same point in a right line (GD) given in position, two right lines whose difference shall be a maximum. Fig. 31.

Draw from A the right line AI at right angles to DG, and produce it until IC is to IA; join BC, and produce it to D; join DA.

Then it is evident that CB is the difference between AD and DB: it is greater than the difference between. any other lines (AO,OB) drawn to any point O in DG; join OC. Then AO is = to OC, but OB is less than OC and CB, .. the difference between BO and OC is less than CB, ... the dif. between AO and OB is less than CB,.. &c.

PROP. 77, THEOR.

If two right angled triangles (ABC,DEF) having equal hypothenuses, (AC,DF,) have two other sides also

equal, the remaining sides shall be equal, and the two triangles shall be equal in every way.

Fig. 55.

Produce AB until BG is = to ED, join GC.

Then the triangles BCG and DEF have BG = ED and BC EF, and the angle CBG FED, they will also have the angle G = D, and GC = DF = AC: whence the triangle ACG being isosceles, the angles G or D will be A, and consequently the angle F = ACB. the triangles ABC and DEF being equiangular, and having AC DF they are in every respect,

PROP. 78, THEOR.

If two triangles (ABC,DEF) have two sides (AC,BC) of the one respectively equal to two sides (DF,EF) of the other, and have the angle (CAB) subtended by the side (CB) equal to the angle (FDE) subtended by the side equal to it (FE); and if the angles (CBA, FED) subtended by the other equal sides, be of the same affection, i. e. both acute or both obtuse; the two triangles will be equal in every respect.

Fig. 56.

=

=

Draw CG,FII at right angles to AB, DE. Then the angle AGC being to DHF, and the angle A D, and the side AC DF, CG is to FH; whence CB being to FF, the angies GBC, HEF are also, (prop. 77 of dedu.) and then the triangles ABC, DEF being mutually equiangular, and having the sides AC and DF = they are in every respect.

If the angles be both obtuse, as in the triangles AIC, DLF, the demonstration is the same:

For if C CB

FE)

FL, the angles GIC, (as before) the angles AIC, DLE shall

PROP. 79, THEOR.

If the angles at the base (ML) of an equilateral triangle be bisected by the right lines, (MI,LI,) the lines