of it, make the adjacent angles ACE, ACB together equal

to two right angles,

therefore BC and CE are in one straight line.

Wherefore, if two triangles &c. Q.E.D.

[I. 14.

PROPOSITION 33. THEOREM.

In equal circles, angles, whether at the centres or at the circumferences, have the same ratio which the arcs on which they stand have to one another; so also have the

sectors.

Let ABC and DEF be equal circles, and let BGC and EHF be angles at their centres, and BAC and EDF angles at their circumferences: as the arc BC is to the arc EF so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF; and so also shall the sector BGC be to the sector EHF.

K

Take any number of arcs CK, KL, each equal to BC, and also any number of arcs FM, MN each equal to EF; and join GK, GL, HM, HN.

Then, because the arcs BC, CK, KL, are all equal, [Constr. the angles BGC, CGK, KGL are also all equal; [III. 27. and therefore whatever multiple the arc BL is of the arc BC, the same multiple is the angle BGL of the angle BGC.

For the same reason, whatever multiple the arc EN is of the arc EF, the same multiple is the angle EHN of the angle EHF

And if the arc BL be equal to the arc EN, the angle BGL is equal to the angle EHN; [III. 27. and if the arc BL be greater than the arc EN, the angle BGL is greater than the angle EHN; and if less, less.

Therefore since there are four magnitudes, the two arcs BC, EF, and the two angles BGC, EHF;

II

and that of the arc BC and of the angle BGC have been taken any equimultiples whatever, namely, the arc BL and the angle BGL;

and of the arc EF and of the angle EHF have been taken any equimultiples whatever, namely, the arc EN and the angle EHN;

and since it has been shewn that if the arc BL be greater than the arc EN, the angle BGL is greater than the angle EHN; and if equal, equal; and if less, less;

therefore as the arc BC is to the arc EF, so is the angle BGC to the angle EHF. [V. Definition 5.

But as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF,

for each is double of each;

[V. 15.

[III. 20.

therefore, as the arc BC is to the arc EF so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Also as the arc BC is to the arc EF, so shall the sector BGC be to the sector EHF.

Join BC, CK, and in the arcs BC, CK take any points X, O, and join BX, XC, CO, OK.

Then, because in the triangles BGC, CGK, the two sides BG, GC are equal to the two sides CG, GK, each to each;

and that they contain equal angles;

[III. 27.

therefore the base BC is equal to the base CK, and the triangle BGC is equal to the triangle CGK.

[I. 4.

A

M

N

And because the arc BC is equal to the arc CK, [Constr. the remaining part when BC is taken from the circumference is equal to the remaining part when CK is taken from the circumference;

therefore the angle BXC is equal to the angle COK. [III. 27. Therefore the segment BXC is similar to the segment COK; [III. Definition 11. and they are on equal straight lines BC, CK. But similar segments of circles on equal straight lines are equal to one another;

[III. 24. therefore the segment BXC is equal to the segment COK.

And the triangle BGC was shewn to be equal to the triangle CGK;

therefore the whole, the sector BGC, is equal to the whole, the sector CGK. [Axiom 2.

For the same reason the sector KGL is equal to each of the sectors BGC, CGK.

In the same manner the sectors EHF, FHM, MHN may be shewn to be equal to one another.

Therefore whatever multiple the arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC;

and for the same reason whatever multiple the arc EN is of the arc EF, the same multiple is the sector EHN of the sector EHF.

And if the arc BL be equal to the arc EN, the sector BGL is equal to the sector EHN;

and if the arc BL be greater than the arc EN, the sector BGL is greater than the sector EHN; and if less, less.

Therefore, since there are four magnitudes, the two arcs BC, EF, and the two sectors BGC, EHF;

A

M

E

and that of the arc BC and of the sector BGC have been taken any equimultiples whatever, namely, the arc BL and the sector BGL;

and of the arc EF and of the sector EHF have been taken any equimultiples whatever, namely, the arc EN and the sector EHN;

and since it has been shewn that if the arc BL be greater than the arc EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less ;

therefore as the arc BC is to the arc EF, so is the sector BGC to the sector EHF. [V. Definition 5.

Wherefore, in equal circles &c. Q.E.D.

PROPOSITION B. THEOREM.

If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD: the rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square on AD.

Describe the circle ACB about the triangle,

[IV. 5.

[III. 21.

therefore the triangle BAD is equiangular to the triangle EAC.

Therefore BA is to AD as EA is to AC;

[VI. 4.

[VI. 16.

therefore the rectangle BA, AC is equal to the rectangle EA, AD,

that is, to the rectangle ED, DA, together with the square on AD.

[II. 3.

[III. 35.

But the rectangle ED, DA is equal to the rectangle BD, DC; therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square on AD.

Wherefore, if the vertical angle &c. Q.E.D.