Conversely:

In equal circles equal arcs are subtended by equal angles.

In the equal circles AFM, BHN let the arcs FMG, HNK be equal.

Then shall the 4s FPG, HQK, which these arcs subtend at the centres, be equal.

For let the

AFM be applied to the

BHN,

so that the centre P may be on Q and PF on QH,
the Oce AFM on the Oce BHN.

and..

Then G will coincide with K, ·.· arc FMG is

.. PG will coincide with QK;

.. LFPG is = L HQK.

= arc HNK.

COR. In equal circles equal arcs are subtended by equal

chords.

For since Fand G coincide with H and K;

.. the chord FG will coincide with the chord HK,

PROBLEM D.

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

Q

Let AB be the given straight line and Q the given 4. It is required to describe upon AB a segment of a circle which shall contain an 4 = Q.

If the Q is a right 4,

Bisect AB in C, and with C as a centre at the distance CA describe a O.

Then segment on AB will be a semicircle,

and will... contain an. <= Q, a right. (III. 16)

Then at the point A in AB make ▲ BAT=

Through A draw AD 1 to AT;
bisect AB in X;

through X draw XC 1 to AB cutting AD in C;

with centre Cat the distance CA describe a .

Join CB.

Then . AX, XC, and ▲ AXC are respectively = BX, XC, and 4 BXC,

.. AC BC (I. 1); .. the passes through B:

=

and the segment AZB (which is alternate to L BAT) shall be the one required.

...AT is to ACD, .. AT touches the ; (III. 6)

... L BAT= 4 in segment AZB;

.. Q4 in segment AZB;

... AZB is the segment required.

(III. 17)

PROBLEM E.

From a given circle cut off a segment which shall contain an angle equal to a given rectilineal angle.