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proportionals between CA and BD (32. cor. 1), viz.; CA to CD as BA to BD and alternately CA to BA as CD to BD. Next, because the triangles BAD and EAC are similar and BA to AD as EA to AC, therefore BA.AC-EA.AD (P. 22). But EA.AD=ED.DA+ square of AD (P. 28). And also ED.DA=BD.DC(P. 37). Therefore BA.AC= BD.DC+square of AD.

Conversely, it can be proved that if BA to AC as BD to DC, or, if, BA.AC=BD.DC+ square of DA. Then DA bisects the angle BAC. Therefore if any angle &c. Q.E.D.

PROP. XL. THEOREM. (E. 3. 18).

If a straight line touch a circle, the straight line drawn from the centre to the point of contact, is perpendicular to the line touching the circle. And conversely, if a straight line passing through an extremity of a diameter is at right angles to it, the straight line touches the circle.

Let the straight line DE touch the circle ABC at C, and from F the centre, let the straight line FC be drawn to C the point of contact. FC is perpendicular to DE.

For, if not, draw FG perpendicular to DE. Because FGC is a right angle, the side FC opposite to it is greater than the side FG (P. 23 and B). But FB is equal to FC (Def. 6.) Therefore FG is greater than FB. This is absurd. Wherefore FC is at right angles to DE.

Conversely, if DE passes though C at right angles to the diameter AC, it touches the circle at C.

Because it can be shown that if any other line than

DE could touch the circle at the point C, the angles made by that straight line at C with the diameter AC would be right angles; which could be shown as absurd, because the angles made by DE at C with the diameter AC are right angles. Therefore DE touches the circle. Wherefore if a straight line &c. Q. E. D.

Cor. It is manifest that a straight line can be drawn touching a circle from a point, cither within its circumference or without it.

PROP. XLI. THEOREM. (E. 3. 22).

The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles and conversely if the opposite angles of a quadrilateral figure are together equal to two right angles, a circle can be described about it.

First let ABCD be a quadrilateral figure inscribed in the circle ABCD. Any two of its opposite angles are together equal to two right angles.

Because the angle ADB is equal to the angle ACB (P. 36) and the angle DCA to the angle DBA (P. 36).

therefore the whole angle DCB is equal to the two angles ADB and DBA. Add the angle DAB to both, therefore the three angles of the triangle ADB are equal to the two opposite angles DCB and DAB. Because the three angles of the triangle DAB are together equal to two right angles (P. 23), therefore the two opposite angles DCB and DAB are also equal to two right angles (Ax. 1). Therefore also the opposite angles ADC and ABC are equal to two right-angles (23. cor.).

Conversely, let ABCL be a quadrilateral figure having its opposite angles LAB, LCB together equal to two right angles. A circle can be described about it.

Bisect AB and BC at G and H. Draw GF and HF at right angles to AB and BC. Join FA, FB, FC.

H

If can be shown that FA, FB and FC are equal (P. 3)· Therefore if from F as centre with any of the straight lines FA, FB and FC, a circle be drawn, its circumferance will pass through the points A,B,C, and also through L. For, if not, let L be without the circle. Take any point D in the arc intercepted between AL and CL. Join AD, CD. Because DABC is a quadrilateral figure inscribed in a circle, therefore the opposite angles DAB and DCB are together equal to two right angles. This is absurd. Therefore the circumference will pass through the point L. And a circle can be described about the quadrilateral figure ABCL. Wherefore the opposite &. Q.E.D.

Lema. Any regular polygon may have one circle described about it and another inscribed in it, and the same point is the centre of both circles.

For, if ABCD be a regular polygon, i.e., have all its sides and angles equal and if any two of the adjacent sides AB and BC be bisected at G and H and from the points G and H perpendiculars GF and HF be drawn and also AF, BF and CF be joined, it can be proved that the sides AF, FB, FC are equal (P. 3), that the angles FAB, FBA, FBC and FCB are equal (P. 3 and 18. cor.) and that each of them is a half of the whole angle. Now if straight lines be drawn from F to all other angles and

also perpendiculars to all other sides of the polygon, these straight lines can be shown to be equal to the first straight lines AF, BF and FC and also the perpendiculars to the first per endiculars. If therefore a circle be described with the point F as centre at the distance of any one of these straight l'nes, it will pass through all the angular points of the polygon, and will be described about it. It is plain next that another circle can be described with the point F as centre at the distance of any of the perpendiculars, and it will pass through the extremities of all the others and will touch the sides of the polygon (P. 40).

PROP. XLII. THEOREM.

If a straight line touch a circle. and from the point of contact a straight line be drawn meeting the circle; the angles which this straight line makes with the line touching the circle shall be egual to the angles which are in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn, meeting the circumference in D, and dividing it into the segments DCB, DAB, of which DCB is less than, and DAB greater than a semicircle.

Then the angles which BD makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle; that is the angle DBF shall be equal to the angle which is in the segment DAB and the

angle DBE shall be equal to the

segment DCB.

angle in the alternate

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right angles to EF, meeting the circumference in A;

Take any point C in the arc DB, and join AD, DC, CB.

E

Because the straight line EF touches the circle ABCD in the point B.

And BA is drawn at right angles to the touching line from the point of contact B (P. 41) the center of the circle is in BA therefore the angle ADB in a semicircle is a right angle (24. cor.) and consequently the other two angles BAD, ABD are equal to a right angle (P. 23). But ABF is likewise a right angle (constr). Therefore the angle ABF is equal to the angles BAD and ABD. Take from those equals ABD, therefore the remaining angle DBF is equal to the angle BAD which is in BDA the alternate segment of the circle. And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD aro equal to two right angles (P. 40). But the angles DBF, DBE are equal to the angles BAD, BCD. And DBF has been pro ved equal to BAD. Therefore the remaining angle DBE is equal to the angle BCD in BDC, the alternate sigment of the circle. Wherefore if a straight line &c. Q. E. D.

Cor. It is plain that a triangle can be inscribed in a circle equiangular to a given triangle.

Because, if ABC be a triangle, let EF a straight line touch the circle ABC, draw BA perpendicular. Place the angle ABC upon the angle ABE and the angle ACB upon the angle ABE. Therefore an equiangular triangle can be inscribed in the circle.

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