Fig. 7.

1st. If DF is not par. to BC draw DG parallel to it, then since AB is bisected in D and DG parallel to the base; AC is bisected in G, but it is also bisected in F which is absurd thus it can be proved that no line but DF can be drawn through the points D or F darallel to BC. 2nd. Since AB is bisected in D the triangle ADF is to BDF and is .. half of ABF, but ABF is half of ABC.. ADF is the fourth part of ABC.

PROP. 8, THEOR.

In a right angled triangle (ABC) the right line joining the point of bisection of hypothenuse, and the right angle (CBA) is equal to half the hypothenuse.

Fig. 8.

ANALYSIS.

Then since AD is to DB

Suppose it to be the case. the angles DBA and DAB are =. Draw DE parallel to BC; then (by prop. 6. of analy.) AE and EB are = and the angles AED and DEB are also =, .. AED is a right angle,.. the triangles AED and DEB are equal in every way,.. AD, DB and DC are to one another.

Cor. Thus it is evident that the angle in à semicircle is a right angle. Since from the centre D a circle might be described (through the three points C, B, A,) of which AC would be a diameter.

Or, it may thus be proved that the angle in a semicircle is a right angle.

are

Since DB and DC are the angles DCB and DBC (5. 1. Elr.) for the same reason the angles DBA and DAB are = .. the angle CBA is to the two BCA and BAC.. (by cor. 1. 32. 1. Elr.) CBA is a right angle.

PROP. 9, PROB.

To draw from two given points (CH) right lines making equal angles at the same point of a line (AB) given in position. Fig. 9.

ANALYSIS.

Suppose it to be done, and that GC, HD are the lines, draw GI at right angles to AB and produce it to meet HD (produced) iu C, then because the angle GDA is =

to HDB it is also to IDC which is vertically opposite to HDB and GID is to CID being both right angles and ID is common to the two triangles DCI, DGI, .'. GI is to IC.

Therefore it is evident if you let fall a perpendicular GL (from either of the points) on the given line and produce it until the produced part IC is to Gl and connect the extremity C with the other given point H the point D where the connecting line cuts the given line is the point required. For, join GD, since GI is to IC and ID common and the angles GID and CID; the angles iDG IDC are (prop. 4. 1. Elr.) but IDC is to HDB (prop .15. 1. Elr.) .. HDB and GDI are =.

PROP. 10. THEOR.

If the right lines (GD, HD) drawn from two given points make equal angles with a right line (AB) given in position they are the shortest lines that can be drawn from those points to any one point in that line.

Fig. 9.

For if possible let GE, and HE be drawn less than them; draw GI at right angles to AB and produce till it meets HD produced in C; it is evident that DC is to DG (4. 1. Elr.) and .. that the whole HC is

to GD and DH; it is also evident that if EC be drawn it is to GE and .. that GE and EH are to HE and EC, but HC is less than HE and EC ... GD and DH are less than GE and EH; in like manner they can be proved less than any other two lines &c.

PROP. 11, PROB.

To find a point in the side (AC) of a triangle (ABC) from which a line may be drawn to another side (AB) parallel to the third side and equal to the segment (of the side to which it is drawn) that is conterminous with the parallel side (BC.)

Because DE is par. to BC the angles EDB and DBC are and because DE is to EB the angles DBE and

BDE are therefore the angles DBC and DBE are. Therefore it is evident if the angle ABC opposite the line in which the point is to be found is bisected by BD that D is the required point. For since DE is par. to BC, angle DBC is to BDE ... BDE is to EBD and ... DE is to EB.

PROP. 12, PROB.

To draw a perpendicular to the extremity (C) of agiven line (AC) without producing it.

Fig. 11.

Take any point B in the given line and from it draw BD at right angles and to CB and through Ddraw DE at right angles to BD, draw BE bisecting the angle CBD to meet DE, join EC; EC is at right angles to AC.

For in the triangles EBD, ECB the sides CB nd BE are respectively to DB and BE and the angle DBE is

to CBE.. the angle BDE is to the angle BCE (4. 1. Elr.) but BDE is a right angle.. BCE is a right angle.

NOTE-DE and BE will meet since the angles DBE and BDE are less than two right angles, (ax. 12)

PROP. 13, PROB.

To draw through a given point (A) placed between two given right lines (CD, CE) that are not parallel a a right line terminated by the given lines and bisected in the given point.

Fig. 12.

ANALYSIS.

Suppose it done; then through A draw AF par. to CB, and AG parallel to CD; because DA is to AB and the angles DAF, ABG as also the angles DFA, AGB (prop. 28. 1. Elr.) the triangles DFA, AGB are in every way.. FA is to GB.

Therefore, it is evident if through the given point, lines AF,AG are drawn parallel to the given ones CD, CE and if GB be taken to CG or FA that the right line BD drawn from B through the given point is terminated by the given lines and bisected in the given point (prop. 26. 1. Elr.)

PROP. 14, THEOR.

If a right line (CD) drawn from the vertex of any tris angle (CAB) bisects the base, it bisects every line parallel to the base.

Fig. 13

For, if it does not, draw CH to bisect EF (par. to the base.) Join ED, FD and HD. Then because EH and HF are the triangles CHE, CHF are= (prop. 38. 1. Elr.) also the triangles EHD, FHD are.. the figures CHOF, CHDE are, but the triangles CD B, CDA are= and also the triangles DFB, EDA, ... CDF and CDE are, then CDF is half CEDF and also CHDF is half CEDF.. CDF and CHDF are which is absurd .. EF is not bisected in H and in like manner it can be proved that it cannot be bisected in any point but G.

PROP. 15, THEOR.

Of all the triangles that can be formed having two sides of the one respectively equal to two sides of the other, the greatest is that which has those two sides at right angles to one another.

Since the triangle DAB, which has the side DB (to - CB) at right angles to the common side AB, is greater than the triangle CAB, the line CD joining their vertices will not be parallel to AB; for if it was the angle CDB would be right, since ABD is right; then CB would be greater than DB but it is to it; .. the line CE drawn from C par. to AB must fall below the point D, then it is evident if you join EA that the triangle EAB which is to CAB (prop. 37. 1. Elr.) is less than DA B.

PROP. 16. THEOR.

Of all the triangles having the same vertical angle and whose bases pass through a given point (A) the least is that (viz. CBD) whose base (BD) is bisected in the given point.

Fig. 12.

Let CEI be any other triangle whose base El passes through the given point, and is not bisected in it; it is greater than CBD.

F

Draw BH parallel to CD; then since BA and AD are =(by hyph.) and the angles BAH, IAD are = (prop. 15. 1. Elr.) and the angles AID, AHB are = (prop. 29. 1. Elr.) the triangles BAH and IAD are = to_one another, to each add the figure BAIC, then BDC and BHIC are.. CEI is greater than CBD.

PROP. 17. THEOR.

If two finite right lines (DE, BC) bisect one another, the right lines (DB, BE, EC, CD) joining their extremities form a parallellogram.

Fig. 15.

ANALYSIS.

Since DBEC is a parallellogram, the triangles DAB and CAE are and the angles AEC and ADB are also and also the joining lines are respectively.

Then since in the triangles DAB, and CAE, the sides DA and AB are respectively to CA and AE, and the angles DAB and CAE are= (prop. 15. 1. Elr.) the triangles themselves are = and the side DB is = to CE and the angle BDE is to DEC.. DB is parallel to CE in like manner it can be proved that DC is par. and to BE... DBEC is a parallellogram.

Cor. It is also evident that if two equal right lines bisect one another at right angles, the right lines joining their extremities form a square.

PROP. 18. THEOR.

If from one side (AB) of a selene triangle (ABC) a part (AD) be cut off equal to another side (AC) and the point of section be connected with the opposite angle; the angle contained by this connecting line and third side, is equal to half the difference of the angles at the third side.

Fig. 16.

Since AD is to AC the angles ADC and ACD are , but the angle ADC is to DBC and BCD ... ACD is = to DBC and BCD.. the whole angle ACB is = to ABC and twice BCD.

=