Fig. 34.

Let O be the centre, join OE, OA, OB and OD. Then in the triangles OAB, OED the sides OA, OB are respectively to OE OD, the angles OAB, OED right angles and the angles ODE, OBA of the same affection, the triangles are in every way (prop. 78 on b. 1.).. BA is to DE.

PROP. 28. THEOR.

The diagonals (BE, AC,) of a regular pentagon cut one another in extreme and mean ratio.

Fig. 35.

In the pentagon ABCDE, the diagonal BE may be proved parallel to DC and.. EII = to EA and .. to AB, and it can also be proved that AC is bisected by the right line BF drawn at right angles to ED, hence the rectangle AGCBG is to AB or EH2, but the rectangle AGC is to the rect. AHC ( to the rect. EHB) hence the rect. EHB+ HB' is to EH i. e. the rect. EBH is to EH... &c.

24

PROP. 29. PROB.

Given the area of a rectangle and the excess of one side above another to construct it.

Suppose that DA and EC are to the sides of the required rectangle, and ED = to the difference, then it is evident that AD is the secant pa sing through the centre of a circle to which the side of a, to the given area is a tangent, drawn from the extremity A. Then let AB be to the area, draw CB at right angles to it to half the difference, from C as a centre with CB as radius describe a circle and draw AD through the centre.

=

It is evident that the rect. DAE is to AB2 and . to the given area.

If two circles touch one another internally(in C,)and if a right line (PR) be drawn cutting both, and right lines (PC, TC, RC, XC) be drawn from the points of section to the point of contact the angle (PCT) contained by one pair of them is equal to the angle (RCX) contained by the other pair.

Fig. 37.

Draw a tangent LM through the point of contact, then the angle LCP is to the angle CRP, and the angle LCT to the angle CXT, but CX'T is

to CRX, XCR, LCT is to them, take away the parts LCP, CRP and the remainders PCT, XCR are .. &c. &c.

PROP. 31. PROB

Draw two tangents to given circle, which shall contain an angle equal to a given rectiliaeal angle. Fig. 38.

ANALYSIS.

Suppose it done and that AB, AD are the required tangents, and that BAD is to to the given angle ; join the points of contact B, D with the centre I. Then because the angles IBA, IDA are right the angles DIB DAB are together to two right angles, .. BID is the supplement of the given angle.

Therefore, draw a diameter ED, and from the centre I draw ID, IB making each of the angle BIE, DIC= to half the given angle, and draw the tangents BA, DA from D, Bit is evident that BAD is to the given angle for BID is the supplement.

PROP. 32. THEOR.

If the adjacent sides of a quadrilateral figure whose opposite angles are together equal to two right angles, be bisected by perpendiculars, and perpendiculars be drawn from their intersection to the other sides, they bisect them.

For, since the opposite angles are equal to two right angles the figure might be inscribed in a circle, .. the point of intersection of the perpendicnlars is the centre, ..all perpendiculars from that point on any lines in the eircle bisect them.

PROP. 33. THEOR.

If any points (A,C) be taken in the circumference of a given circle, and right lines be drawn from these to a point in a tangent, those drawn from them to the point of contact (D) contain u greater angle than those to any other point (E).

Fig. 39.

Join AB, then it is evident that the angle ABC is greater than BEA, but ABC is ADC.. &c.

PROP. 34, PROB.

Given the segments of the base made by a right line bisecting the vertical angle, and the vertical angle to find the triangle.

Suppose that EAB is the required triangle and AC, CB the given segments; describe about it a circle, join EC and produce it to D.

Then since the angles AED, BED are the segments AD, DB on which they stand are equal.

Then describe on (AB) the sum of the segts. a segment of a circle capable of containing the given angle complete the circle and bisect the part of it AB in D, join DC and produce it to E, join AE, BE; then AEB is evidently the required triangle.

PROP. 35. THEOR.

If two tangents (AB,AD) be drawn from the same point to a given circle and the points of contact (BD) be joined, the angle contained by the tangents is equal to the difference of the angles in the segments.

Fig. 38.

For the angle SBD is but the angle SBD is

to the angles ABD and BAD ; to the angle BGD, and the

angle ABD is = to the angle BFD ... the angle BGD is to the angles BAD, BFD.. BAD is the difference between BGD and BFD.

Y

1

PROP. 36. THEOR

If a right line (BF) be drawn from an angle of a regular pentagon bisecting the opposite side it passes through the centre of the circumscribin circle.

Fig. 35.

For the right lines BE BD are evidently to one another since BF bisects ED it is at right angles to it.. it passes through the centre.

PROP. 37. THEOR.

If two right lines (AB, CB) make an angle, and be cut so that the rectangle under the whole lines and segments next the angles be equal, the circle described to pass through their extremities and one point of section (D) shall also pass through the other point of section.

Fig. 41.

Draw from the point B a tangent BG; then if the circle does not pass through the point of section E let it pass through F; then the rectangles ABF and CBD. are each to BG' (prop. 36. b. 3. Elr.) ... they are to another but the rect. ABE is to the rect. CBD (by hypoth.).. the rect. ABF is to the rect. ABE the greater to the less which is absurd.

PROP. 38. THEOR.

If two chords (AB CD) cut one another within a given circle, the angle of their inclination is equal to half the angle at the centre which stands on an arch equal to the sum of the arches intercepted between them.

And if they cut one another without the circle, the angle of their inclination is equal to half the angle at the centre which stands on the difference of the arches which they subtend.

Fig. 42.

1. Draw BL parallel to DE, let I be the centre, then AL is to the sum of the arches AC DB, and the angle of inclination AOC is to the angle ABL (prop. 28. b. 1. Elr.) but the angle AIL at the centre is double of the angle ABL.. the angle of inclination, AOC is = to half the angle AIL.

Fig. 43.

2. If they meet without the circle, draw DL parallel to BA, let I be the centre, join CI LI; then the angle CDL is half of CIL, but CDL is ≈ to COA .. COA is to half CIL.

PROP. 39 THEOR.

The vertical angle (ELF) of any triangle (ELF) inscribed in a circle is greater or less than a right angle, by the angle contained by the side subtending it and the diameter drawn from the extremity of this side.

Fig. 44.

If the given angle FLE be obtuse, join AL; then the angle ALE is to the angle AFE, and FLA is a right angle .. &c.

And if the angle FLE be acute, draw the diameter EC, join CL.

Then the angles CLF CEF are, but CLE is a right angle ... &c,

PROP. 40. PROB.

Three points (A,B,C) being given, in the circumference of a given circle, equally distant from one another to describe two equal circles which shall touch one another in the middle point, and pass one through one of the extreme points, and the other through the other.

Fig. 45.

Draw through the middle point B a tangant, join BA BC and make the angles BCI BAO each respectively = to the angles ABO CBI, the points O and I are evidently the centres of the required circles.