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

D

N

M

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;

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.

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.

PROPOSITION C. THEOREM.

If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a triangle, and let AD be the perpendicular from the angle A to the base BC: the rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle. Describe the circle ACB

therefore the triangle ABD is equiangular to the triangle AEC.

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

PROPOSITION D. THEOREM,

The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle is equal to both the rectangles contained by its opposite sides.

Let ABCD be any quadrilateral figure inscribed in a circle, and join AC, BD: the rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD and by AD, BC.

Make the angle ABE equal to the angle DBC; [I. 23.

[VI. 16.

therefore the rectangle AD, CB is equal to the rectangle

DB, EC.

Again, because the angle ABE is equal to the angle DBC, [Construction. and the angle BAE is equal to the angle BDC, for they are in the same segment of the circle; [III. 21. therefore the triangle ABE is equiangular to the triangle DBC.

Therefore BA is to AE as BD is to DC;

[VI. 4.

therefore the rectangle BA, DC is equal to the rectangle AE, BD.

[VI. 16.

But the rectangle AD, CB has been shewn equal to the rectangle DB, EC;

therefore the rectangles AD, CB and BA, DC are together equal to the rectangles BD, EC and BD, AE;

that is, to the rectangle BD, AC.

Wherefore, the rectangle contained &c. Q.E.D.

[II. 1.