AD=CD, and DB is common, and ▲ ADB = ▲ CDB,

But, in the same circle, the arcs, which are subtended by equal chords, are equal, the greater to the greater and the less to the less;

and BD, if produced, is a diameter,

III. 28.

.. each of the arcs BA, BC, is less than a semicircle,

and .. arc BA=arc BC.

Thus the arc ABC is bisected in B.

Q. E. F.

Ex. If, from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, and the other at right angles to the diameter, the squares on these two lines are together double of the square on the radius.

PROPOSITION XXXI. THEOREM.

In a circle, the angle in a semicircle is a right angle; and the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

B

Let ABC be a O, O its centre, and BC a diameter.
Draw AC, dividing the into the segments ABC, ADC.

Join BA, AD, DC, AO.

L

Then must the ▲ in the semicircle BAC be a rt. ▲, and ▲ in segment ABC, greater than a semicircle, less than a rt. L, and in segment ADC, less than a semicircle, greater than a rt. 4 . BCAO, .. ▲ BAO= 4 ABO;

First,

.. 4 COA=twice ▲ BAO;

and : CO=A0, .. L CAO= L ACO ;

.. BOA=twice ▲ CAO;

I. A.

I. 32.

I. A.

I. 32.

.. sum of 4 s COA, BOA=twice sum of ▲ 8 BAO, CAO, that

is, two right angles = twice / BAC.

.. BAC is a right angle.

Next, BAC is a rt. 4,

.. ABC is less than a rt. 4.

Lastly, sum of 4 s ABC, ADC=two rt. 48,

and

ABC is less than a rt. 4,

.. 4 ADC is greater than a rt. 2. NOTE.-For a simpler proof see page 178.

I. 17.

III. 22

Q. E. D.

Ex. 1. If a circle be described on the radius of another circle as diameter, any straight line, drawn from the point, where they meet, to the outer circumference, is bisected by the interior one.

Ex. 2. If a straight line be drawn to touch a circle, and be parallel to a chord, the point of contact will be the middle point of the arc cut off by the chord.

Ex. 3. If, from any point without a circle, lines be drawn touching it, the angle contained by the tangents is double of the angle contained by the line joining the points of contact, and the diameter drawn through one of them.

Ex. 4. The vertical angle of any oblique-angled triangle inscribed in a circle is greater or less than a right angle, by the angle contained by the base and the diameter drawn from the extremity of the base.

Ex. 5. If, from the extremities of any diameter of a given circle, perpendiculars be drawn to any chord of the circle that is not parallel to the diameter, the less perpendicular shall be equal to that segment of the greater, which is contained between the circumference and the chord.

Ex. 6. If two circles cut one another, and from either point of intersection diameters be drawn, the extremities of these diameters and the other point of intersection lie in the same straight line.

Ex. 7. Draw a straight line cutting two concentric circles, so that the part of it which is intercepted by the circumference of the greater may be twice the part intercepted by the circumference of the less.

If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle must be equal to the angles, which are in the alternate segments of the circle.

Let the st. line AB touch the CDEF in F.

Draw the chord FD, dividing the

Then must

into segments FCD, FED. DFB= 4 in segment FCD,

and DFA in segment FED.

=

From F draw the chord FC1 to AB.

Then FC is a diameter of the .

Take any pt. E in the arc FED, and join FE, ED, DC.

Then ·· FDC is a semicircle, .. ▲ FDC is a rt. 4;

.. sum of 8 FCD, CFD=a rt. 4.
Also, sum of 4 s DFB, CFD=a rt. 4.

.. sum of 4 s DFB, CFD=sum of ▲ s FCD, CFD,
and .. 4 DFB= ▲ FCD,

that is, ▲ DFB= ▲ in segment FCD.

III. 19.

III. 31.
I. 32.

III. 22.

Again, . CDEF is a quadrilateral fig. inscribed in a ©,
.. sum of 4 s FED, FCD=two rt. 4 s.
Also, sum of 4 s DFA, DFB=two rt. 4 s.

.. sum of 2 s DFA, DFB=sum of 2 s FED, FCD ;
and DFB has been proved ▲ FCD ;

=

.. 4 DFA= = ▲ FED,

that is, ▲ DFA = in segment FED.

I. 13.

Q. E. D.

Ex. The chord joining the points of contact of parallel tan

gents is a diameter.

PROPOSITION XXXIII.

PROBLEM.

On a given straight line to describe a segment of a circl capable of containing an angle equal to a given angle.

Let AB be the given st. line, and C the given 4. It is required to describe on AB a segment of a which shall contain an ▲ = L C.

I. 23.

At pt. A in st. line AB make ▲ BAD= t C. Draw AEL to AD, and bisect AB in F. From F draw FG1 to AB, meeting AE in G. Join GB. Then in As AGF, BGF;

... AF=BF, and FG is common, and ▲ AFG= ▲ BFG ;

.. GA=GB.

L

With Gas centre and GA as radius describe a

Then will AHB be the segment reqd.

I. 4.

ABH.

For AD is 1 to AE, a line passing through the centre,
.. AD is a tangent to the ABH. III. 16.
And the chord AB is drawn from the pt. of contact A,
.. < BAD= 4 in segment AHB,

that is, the segment AHB contains an = 4 C,
and it is described on AB, as was reqd.

III. 32.

Q. E. F.

Ex. 1. Two circles intersect in A, and through A is drawn a straight line meeting the circles again in P, Q. Prove that the angle between the tangents at P and Q is equal to the angle between the tangents at A.

Ex. 2. From two given points on the same side of a straight line, given in position, draw two straight lines which shall contain a given angle, and be terminated in the given line.