angle FDB (P. 4). Because BG is equal to GD (Def. 6), therefore the angles GBD and GDB are equal. This is impossible, because the angle GBD has been proved to be equal to the angle FDB which is but a part of GDB.

Conversely, if AC passing through F, the centre, cuts BD at right angles, BD is bisected at E the centre, Join FD, FB.

Because FB and FD are equal (Def. 6), therefore the angle FBE is equal to the angle FDE (Prop. 4). Because of the two triangles FBE and FDE, FB is equal to FD, the angle FBE equal to the angle FDE and the angle FEB to the angle FED, therefore BE is equal to ED (Prop. 10). Therefore BD is bisected at E. Wherefore

if &c. Q. E. D.

Cor. From this it is manifest that the centre of any circle can be found by bisecting at right angles any straight line taken within the circle.

PROP. XXXVI. THEOREM. (E. 3. 20, 21).

The angle at the centre of a circle is double of the angle at the circumference standing upon the same arc, that is, upon the same part of the circumference. And the angles in the same segment of a circle are equal to one another.

Let ABC be a circle, BEC an angle at the centre, and BAC an angle at the circumference,

standing upon the same arc or part of the circumference BC. The angle BEC is double of the angle BAC.

Join AE, and produce it to F. First let the centre of the circle be within the angle BAC.

Because EA is equal to EB, the angle EAB is equal

(I. 4) to the angle EBA. Therefore the two angles EBA and EAB are double of the angle EAB. But the angle BEF is equal (I. 23) to the two angles EAB and EBA. Therefore also the angle BEF is double of the ángle EAB. For the same reason, the angle FEC is double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAC.

Next, let the centre of the ciricle be without the angle BAC.

It may be demonstrated, as in the preceding case, that the angle FEC is double of the angle FAC, and that FEB, a part of the first, is double of FAB, a part of the other. Therefore the re

maining angle BEC is double (III. Ax. 3) of the remaining angle BAC.

Thirdly, Let ABCD be a circle, and BAD and BED angles in the same segment BAED. The angles BAD and BED are equal to one another. First, let the segment BAED be greater than a semicircle.

Let F be the centre of the circle ABCD, and join BF and FD.

Because the angle BFD at the centre, and the angle BAD at the circumference, stand upon the same arc BCD, the angle BFD is double (1st case) of the

angle BAD. For the same reason, the

angle BFD is double of the angle BED. Therefore the angle BAD is equal to the angle BED.

Next, let the segment BAED be not greater than a semicircle.

Join AF, produce it to C, and join CE. Because the segment ADCB is greater B than a semicircle; the angles BAC and BEC are equal, by the first case.

Be

cause CBED is greater than a semicir

cle, the angles CAD and CED are also equal, by the first case. Therefore the whole angle BAD is equal (Ax. 2) to the whole angle BED. Wherefore the angles in the same segment, &c. Q. E. D.

PROP. XXXVII. THEOREM. (E. 3. 35).

If two straight lines cut one another in a circle, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

Let the straight lines AB and CD cut one another at F in the same circle. The rectangle AF.FB

is equal to the rectangle FD.FC. Join AC and BD.

Because the angle AFC is equal to the angle DFB (P. 2) and the angle FAC to

the angle FDB (P. 36). Therefore the remaining angle ACF is equal to the remaining angle DBF (P. 23). Therefore the triangles CAF and BFD are equiangular and similar (P. 18), viz., AF is to FC as FD to FB and therefore the rectangle AF.FB is equal to the rectangle CF.FD (P. 22). Wherefore if two straight lines &c. Q.E.D.

- PROP. XXXVIII. PROBLEM. (E. 4. 5).

To describe a circle about a given triangle.

Let ABC be a triangle. It is required to describe a circle about it. Bisect AB and AC in the points D and E

and from these points draw DF and EF at right angles to

AB and AC respectively (P. 5). The straight lines DF and EF produced meet one another. For, if DE were joined, the angles which DE would make with them would be both acute angles. It is plain, therefore, that they meet one another. Let them meet in F, and join FA. Also, if the point F be not in BC, join BF and CG.

Because AD is equal to DB, and DF which is common to the two triangles ADF and BDF, is at right angles to AB, the base AF is equal (P. 3) to the base FB. In like manner, it may be shown that CF is equal to FA. Therefore BF is equal (Ax. 1) to FC; and FA, FB, and FC, are equa to one another. Wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q.E.F.

Cor. It is manifest, that when the centre of the circle fall, within the triangle, each of its angles is less than a right angles each of them being in a segment greater than a semicircle; but, when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle (24. cor.). Conversely, if the given triangle be acute-angled, the centre of the

circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it bo an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

PROP. XXXIX. THEOREM. (E. 6. 3).

If any angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base have to one another the same ratio which the adjacent sides have and the rectangle contained by the sides is equal to the rectangle contained by the segments together with the square of the straight line bisecting the angle. Conversely, if the sey-ments of the base have the same ratio as the sides or the rectangle contained by the sides is equal to the rectangle contained by the segments with the square of the straight line, the straight line bisects the angle.

Let ABC be a triangle and the angle BAC bisected by AD. First BA to AC as BD to DC.

Describe a circle about ABC (P. 38).

Produce AD to E. Join EC.

D

Because in the two triangles ABD and CED, the angle ADB is equal to the angle CDE (P. 3), the angle at B to the angle at E (P. 36) and the angle ECD to the angle BAD (P. 36), therefore EC to CD as BA to AD. Because in the two triangles ABD and AEC, the angle ABD is equal to the angle AEC (P. 36), the angle BAD to the angle EAC (hyp.) and consequently the angle ADB to the angle ACE (P. 23), therefore EC to CA as BD to AD. But it has been shown that EC to CD as BA to AD. Therefore CD and BA are mean