THEOREM (e).

If from a point without a circle a secant and a tangent be drawn, the rectangle contained by the parts of the secant between the given point and the circle is equal to the square on the tangent.

From a point A without the ODTR let the tangent AT and the secant ADR be drawn.

Then shall rectangle (AR, AD) = square on AT.

Ist. If the secant passes through the centre C: join CT.

=

D

Then ... AR is the sum, and AD= the difference of CD and CA;

.. rectangle(AR,AD) + square on CD = square on CA, (II. 2) and.. = squares on CT, AT; (111. 6), (1. 35)

but square on CD=square on CT;

.. rectangle (AR, AD) = square on AT.

Next: If the secant does not pass

through the centre,

let fall CPL on ADR,

and join CD, CT, CA.

Then rectangle (AR, AD) + square

on DP = square on AP;

R

(II. 2)

.. rectangle (AR, AD) + squares on DP, CP= squares on AP, CP;

=

... rectangle (AR, AD) + square on CD square on AC, squares on CT, AT; (I. 35), (III. 6)

and ...

=

but square on CD = square on CT;
.. rectangle (AR, AD) = square on AT.

COROLLARY. If from any point A without a TDG, © two straight lines ADR, AGS be drawn cutting the ©, then rectangle (AD, AR) is = rectangle (AG, AS).

DG

For each of them is the square on the tangent AT drawn from A to the ..

Conversely :

If from a point A without a two straight lines are drawn, one of them ASR cutting the, and the other AQ meeting it, and rectangle AS, AR = square on AQ: then shall AQ touch the.

P

R

A

For if not, if possible, let it cut the in some other point P.

Then rectangle AQ, AP= rectangle AS, AR, and
...=square on AQ, which is impossible;

.. AQ does touch the .

PROBLEM (ƒ).

Find the locus of the vertices of all triangles, on the same base and on the same side of it, having a given vertical angle.

Let AB be the given base, and Q the given angle.

On AB describe a segment of a

ing an angle = Q.

capable of contain

The arc of this segment shall be the locus required.

For join A, B with any point P in the arc :
then the ▲ APB has its vertical angle = Q;

so every ▲ whose vertex is in the arc APB,
having AB for its base, has its vertical ▲ =

Q;

and conversely, every ▲ having its vertical = Q, and AB as its base, and situated on the same side of AB as the segment APB, will have its vertex in the arc APB.

For if not, if possible let AXB be such a ▲ having its vertex within the segment; produce AX to H, join BH. Then ▲ AXB > ▲ AHB, and ... >Q, but it is also = it, which is impossible.

Again, if possible, let ABR be a ▲ on base AB, having its vertical = l,

the vertex R being without the segment; join BK. Then ARB is < AKB, and .. <Q: but it is which is impossible.

also = it;

Hence the locus required is the arc APB.

DEFINITION.

The angle of a segment is that which is contained by the base of the segment and the tangent at one extremity.

THEOREM (g).

Assuming the tangent to be the limiting position of the secant, prove that the angle of a segment of a circle and the angle in it are together equal to two right angles.

Let AT be the tangent at one extremity of the
base AX of the segment ACX.

Then shall the

XAT and the in the segment ACX

be together equal to two right angles.

Through A draw a secant ACR,
and join XC.

Then angles XAC, AXC, and the angle in the segment A CX, are together equal to two right angles. (1. 24), (III. 14) Now let the secant ACR be turned about A

towards AT, and ultimately coincide with it,
then will C move along the arc towards A, and
ultimately coincide with it;

also AXC will continually diminish,
and ultimately vanish.

Hence the 4 XA Tand the 4 in the segment ACX
are together equal to two right angles.

DEFINITION.

Similar segments are those which contain equal

angles.

THEOREM (h).

Similar segments of circles having equal chords are equal in all respects.

A

Р

B C

R

Let APB, CQD be similar segments of circles

having equal chords AB, CD.

Then shall the segments APB, CQD be equal in all respects.

For let the segment APB be applied to the segment CQD so that the chord AB may be on CD;

then will the arc APB be on the arc CQD;

for if not, if possible, let it fall otherwise.

Then since these arcs cannot meet in more points than two, one of them, as CQD, must lie within the other. (111.8) Draw the straight line CQR cutting the arcs in Q and R ; join DQ, DR.

Now.. the segments CQD, CRD are similar

..

LCQD = CRD, which is impossible;

..the arc APB will fall on the arc CQD;

(1. 13)

.. the segments APB, CQD are equal in all respects.

COR.

Upon the same base and on the same side of it there cannot be two similar segments of circles not coinciding with one another.