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

/ FAC,

10 def.

.. each of the s BAC, FAC, is a rt. ▲, BAC in the semi Ort. L.

Wherefore the L, &c.

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 s

which BD makes with the touching line EF shall be equal to the s in the alternate segments of the O; that is,

seg. DAB; and

DBE

=

DBF = in seg. DCB.

in

A

D

Prop. 54.

Prop. 59. ..
Prop. 31.

Then EF touches in B,
and BA EF at the point of contact B;
.. Cr. of is in str. line BA;
ADB in semi ○ =rt. ▲,
< BAD+ABD = rt. ≤;

LABFrt. L,

away

rem.

Prop. 57. ..

▲ ABF = / BAD+ ▲ ABD;
the common ABD,
DBF = frem. ▲ BAD, in al-
tern. seg. DAB.

Again ABCD is a quadrilat. fig. in a ;

Prop. 12. but

BAD+BCD = 2 rt. ▲ S,

DBF+/ DBE = 2 rt. ▲s,

:: <DBF+DBE
and it has been

=

Z BAD+BCD;

proved DBF=<BAD,

.. rem.

<DBE ={rem, BCD, in the

altern. seg. DCB.

Wherefore if a str. line, &c.

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 ; it is required to descr. on AB, a seg. of O, that shall contain an

1st. If C be a rt. .

C.

From Cr. F, and dist. FB, descr. semi

AG = GB,

descr. from Cr. G, dist. GA, passes through B.

Then

AD

Cor. Prop. 51.

Prop. 60.

Const.

Let this

be AHB.

AE at the extrem. of diam. AE, .. AD touches the ;

And AB from pt. of contact, cuts

DAB = in alt. seg. AHB,
DAB = /C,

C in alt. seg. AHB

=

Wherefore, upon the line AB, the seg. АНВ of a is described which contains an ▲ = given

C.

PROP. LXII. THEOR. 35. 3 Eu.

If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other.

Let the two str. lines AC, BD cut each other in E within ABCD: then AE. EC= BE.ED.

1st. Let AC, BD pass each through the Cr. E. Fig. 1. :: AE=EC=BE=ED,

.. AE. EC-BE. ED.

15 def.

2ndly. Let one of them BD pass through the Cr., and cut the other AC which does not pass through the Cr. at rt. s in the point E. Fig. 2.

:: BD which passes through the Cr. cuts AC

which does not pass through the Cr. at rt.

Prop. 49.

s in E,

.. AE = EC,

Prop. 50.

:: BD is cut into equal parts in F, and in 2 unequal parts in E,

.. BE. ED+EF2 = FB2 or FA2,

Prop. 44.

but

AE2+ EFFA',

Prop. 39.

.. BE. ED+ EF2 = AE2+EF2, taking away EF which is common,

BE. ED = AE or AE . EC.

3rdly. Let BD which passes through the
Cr. cut AC which does not pass through the
Cr. in E, but not at rt.
Then as before if BD be bisected in F, F is
Cr. of O.

s.

Fig. 3.