Cor. It is also evident, that if AC be bisected in G, that the triangle DGE is similar to ABC.

PROP. 42, PROB.

To inscribe a square in an equilateral triangle (ABG).

Suppose CF to be the required, let fall a perpendicular BD, join BE. Then LF is to FB, (since the triangle LBF is evidently equilateral,). FB is to FE, the angle FEB is to FBE, but since FE is parallel to BD the angle EBD is to FEB, and .. to FBE.

Therefore it is evident if the perpendicular BD be drawn and the angle DBG be bisected by BE, that the perpendicular raised from the point E to meet BG is the side of the required. For the angle FEB is to EBD and.. to EBF, .. FE is to FB and .. to FL, .'. &c, &c.

PROP. 43. PROB.

Given (AB) the difference between the hypothenuse and

side of an isosceles right angled triangle to con- + Same as 101

struct it,

Suppose ACD to be the required triangle; join CB. Then because AB is the difference, DB is to DC.. the angle DCB is to DBC, but these angles are together the supplement of half a right angle, .. DBC is half the supplement of half a right angle.

2

Therefore, if at one extremity A of the given dif. you make BAC = a right angle, and at the other extremity B, the angle DBC= the supplement of half a right. angle, produce AC and BC to meet, and draw CD at right angles to AC, the triangle ACD is evidently the required one.

PROP. 44, THEOR.

If two triangles (BAC,CAD) have two sides (BA,AC) of the one respectively equal to two sides (CA,AD) of the other, and the contained angles (BAC,CAD) supplemental, the two triangles are equal.

Fig. 38:

For let the

other two

sides BA, AD be placed in directum, the sides must coincide since the angles are supplemental, .. BAC and CAD are of the same altitude, and being on bases are to one another.

PROP. 45. PROB.

To cut from a given line (AB) a part whose square shall be equal to half the square of the given line.

Fig. 39

2

ANALYSIS.

2

to half the AF and draw GE

2

Suppose the of CB to be formed on the given line; bisect AB in G, parallel to BF or AD; GEFB is half the is.. to the □2 of CB, but GF is also HB, the side of an isosceles right angled triangle, formed on the given line, .. CB is to HB.

PROP. 46. THEOR.

2 of AB, and to the of

If any two points (B,C) be taken in the base (DE) of an isosceles triangle (DAE,) the perpendiculars (BF,BG) from one point, are equal to those (CI,CH) from the other point.

Fig. 40.

For draw BK parallel to DA, and CL par. to EA, it is evident that of the two BF and BG, BF is to KI and LG to CH (of the other two CI,CH,) and because BK is par. to DA and CL to AE, the angles BKC and CLB are to one another, for the same reason the angles. KBC and LCB are, and BC common to the triangles BKC and CLB, .. BL is = to BK (by 26. 1. Elr.) .. C1 and CH are together to BA and BG.

PROP. 47, THEOR.

Of all the lines that can be drawn from any point (A) to a given right line (BC) a perpendicular is the least, and of all the others that which is nearest to the perpendicular, is less than the one

mote.

- Fig. 41.

re

For if it be possible, let, AF be drawn less than the perpendicular AD; then because AF is less than AD, the angle AFD is greater than ADF, an acute angle, greater than a right, which is absurd, .. AF is not less than AD. Also, AB is greater than AF; because it subtends a greater angle.

4

PROP. 48, PROB.

To divide a given right line (AB) so that the square of one part shall be equal to twice the square of the other

Suppose that the of AC is to twice the of CB, it is evident that CB must be the side of an isosceles right angled triangle, of which AC is the hypothenuse. Describe on AC an isosceles right angled triangle CDA, join DB. Then each of the angles DAC,DCA is half a right angle, and DCA is to the angles CBD,DCB, but those angles are to one another, because CA and CD are,.. each of them is the fourth part of a right angle.

= =

L

2

Therefore it is evident, if at either extremity A the angle BAD is made to a right angle, and from the other extremity B, BD be drawn to meet it, making the angle ABD of a right angle, and if D be drawn, making the angle BDC to the angle CBD, that C is the required point.

PROP. 49, PROB.

Given the periphery (BE) of an isosceles right angled triangle to construct it.

Suppose it done, and that ADC is the required triangle; join AB and AE, then the angle DCA is = ÷ a : } right angle, .. because CA is to CB the angles CAB and CBA are and each of a right angle; for the same reason each of the angles AED,EAD is of a right angle.

Therefore, draw AE and AB, making each of the angles AEB,ABE to of a right angle, and also draw DA and CA, making the angles DAE and CAB to the angles AEB,ABE, it is evident that the tri. angle ADC is the required one.

PROP. 50, THEOR.

The angle (BAD) of a regular pentagon (EBADC) is trisected by the diagonals (EA,CA).

Fig. 45.

Because the sides EB and BA are to CD and DA, and the contained angles the angles BAE and DAC are; also the angles DAE and CEA are =, and the angles ECD and ADC are,.. EA is parallel to CD, for which reason the angles EAC and ACD are =, but ACD is to CAD, ... EAC is to CAD, ... the entire angle BAD is trisected by the diagonals.

PROP. 51, THEOR.

If from any point (G) in the diagonal of a parallelogram (DABC), right lines be drawn to the opposite angles, they cut off equal triangles.

Fig. 22.

Draw the diagonal DB: then, since the diagonals bisect one another, the triangles DAI and BAI are =, for the same reason the triangles BGI and DGI are ➡, ..the triangles BAG and DAG are,

"

PROP. 52, THEOR.

points (D,E,F,G) be taken in the sides of a square, (ABIC) at equal distances from the angles, the lines connecting them form a square.

Fig. 44.

In the triangles DGC,GFI the sides DC and CG are respectively to GI and IF, and the contained angles are as being right,.. the angle GFI is to DGC, ... the angles DGC and FGI are to the angles FGI, GFI and to a right angle, .. DGF is a right angle: for the same reason the angles GDE,DEF and EFG are right angles, .. the figure is rectangular. Also the sides DG and GF are, and their opposite sides are to them, ... &c.

PROP. 53. THEOR.

If the sides of a regular pentagon be produced to meet in the points (K,L,M,N,O,) all the angles formed at these points are together equal to two right angles.

Fig. 45.

to one

The angles at those points are evidently another: the angle DAB is gths of a right angle, .'. each of the angles KAD,KDA is ths of a right angle, .'. each of the angles K,L,M,N,O, is ths of a right angle, if then they be added together, they are to two right angles.

PROP. 54, PROB.

Given in any triangle the perpendicular let fall from the vertical angle on the base, and the difference between each segment made by the perpendicular and its adjacent side, to construct the triangle.