Let ABCD be a quadrilat. fig. in O ABCD: then any two of its opp. Zs shall be together equal to 2 rt. Zs.

B

Then :: {ZBCA = 2 ADB

Prop. 56.

Join AC, BD.

,

,
Z CAB+ / BCA = whole _ ADC;

to each add _ABC, .. LABC+ 2 CAB+

= _ABC+ LADC,

_ BCA but _ ABC + 2 CAB+/

BCAT

= 2 rt. Ls, .. LABC + LADC= 2 rt. Zs. In like manner it may be shown that

_ BAD+ LDCB= 2 rt. Ls. Therefore, the opp. Ls, &c.

Prop. 31.

PROP. LVIII. THEOR. 26. 3 Eu.

In the same or equal circles, equal angles at

the centre of the circumference stand upon equal arcs.

Let ABG, CDH be equal Os, and the equal angles AEB, CFD at their centres, and

AGB, CHD at their circumferences; then shall arc AB = arc CD.

For, let str. line AE be applied to str. line CF, so that pt. A may coincide with pt. C, and AE fall upon CF.

Then : AE=CF
. pt. E coincides with pt. F.

AE coincides with CF,
And :::

LAEB Z CFD,

EB falls upon FD.
And :::

EB = FD .. pt. B coincides with pt. D; Also :: every point in arc AB is the same dist. from E, as every point in arc CD, is from F; and the points A, B, coincide with points C, D,

arc AB coincides with arc CD,

arc AB = arc CD; .:. the equal _ s at the Cr. AEB, CFD, as also the Ls at the Oce AGB, CHD which are also equal (Ax. 7) stand upon equal arcs AB and CD.

Wherefore, in the same, &c.

i. e.

PROP. LIX. THEOR.
The angle in a semicircle is a right angle.

Let BADC be a semi O, of which the diam. is BC and centre E; also BAC an L contained in the semi O.

Then _ BAC=rt. L

F

D

B

Prop. 5.

And ::

Prop. 5.

Join EA and prod. BA to F. Then, :: BE EA,

LEAB = LABC.

AE EC,

LEAC= LACB,
:. whole _ BAC = L ABC+ LACB;
but, ext. ZFAC = LABC+_ ACB,

_BAC = _FAC,
:: each of the Zs BAC, FAC, is a rt. L,

i. _ BAC in the semi O = rt. L. Wherefore the L, &c.

Prop. 31.

10 def.

PROP. LX. THEOR.

32. 3 Eu.

If a straight line touch a circle, and from

the point of contact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle, shall be equal to the angles which are in the alternate segments of the circle.

Let str. line EF touch O ABCD in B; from B draw BD, cutting the O: then the Ls

which BD makes with the touching line EF shall be equal to the 2s in the alternate segments of the O; that is, ZDBF = L in seg. DAB; and ZDBE = L in seg. DCB.

А

B

Draw BA I EF.
Take any point C in Oce DB;

join AD, DC, CB.

Prop. 54.

Then : EF touches O in B, and BA | EF at the point of contact B;

:. Cr. of O is in str. line BA; Prop. 59. :LADB in semi O =rt. L, Prop. 31. .. _ BAD+ LABD = rt. L; but

LABF =rt. Le ..

LABF=_BAD+ LABD; take away the common _ ABD, rem. DBF

(rem. Z BAD, in al

tern. seg. DAB.

Const.

Ayain :: ABCD is a quadrilat. fig. in a ; Prop. 57. .. _BAD + LBCD

= 2 rt. Ls, Prop. 12. but _ DBF+_ DBE = 2 rt. <s,

.: _DBF+ LDBE Z BAD+ LBCD;
and it has been
proved DBF

:_BAD,

Srem. ZBCD, in the .. rem. LDBE

altern, seg. DCB. Wherefore if a str. line, &c.

{rem

PROP. LXI. PROB.

33. 3 Eu.

Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.

Let AB be a given str. line, C the given L; it is required to descr. on AB, a seg. of O, that shall contain an L=LC.

Ist. If _ C be a rt. L.

Bisect AB in F.
From Cr. F, and dist. FB, descr. semi O

AHB;
Then, any LAHB in semi 0= L C.
2ndly. If _C be not a rt. L.

Prop. 59.

H