The elements of Euclid, explained and demonstrated in a new and most easie method with the uses of each proposition in all the parts of the mathematicks / by Claude Francois Milliet D'Chales, a Jesuit ; done out of French, corrected and augmented, and illustrated with nine copper plates, and the effigies of Euclid, by Reeve Williams ...

PROPOSITION VIII. THEOREM.

IF from a Point without a Circle, be drawn several Lines to its Circum∣ference; first, of all those which are drawn ••o its concave Circumference, the greatest ••asseth through the Center; secondly, those ••earest thereto are greater than those ••hich are farther off; thirdly, amongst ••he Lines which fall on the convex Cir∣••umference, the least being continued, ••asseth through the Center; fourthly, the ••earest thereto are least; fifthly, there ••annot be drawn but only two equal, whe∣••her they be drawn to the convex Cir∣••umference, or that they fall on the con∣••ave.

Let there be drawn from the Point A, several Lines to the Circumference ••f the Circle GC, DE. In the first ••lace the Line AC, which passeth ••hrough the Center B, is the greatest ••f all those which fall on the concave Circumference: for Example, it is great∣••r than AD. Draw the Line BD.

Demonstration. In the Triangle ABD ••he Sides AB, BD, are greater than the

Side AD; now the Sides AB, BC, are equal to AB, BD; thence AB, BC, or AC, is greater than AD.

2. AD is greater than AE.

Demonstration. The Triangles ABD, ABE, have the Side AB common, and the Sides BE, BD, equal; and the Angle ABD, is greater than the Angle ABE: thence (by the 24^th. of the 1^st.) the Base AD is greater than the Base AE.

3. AF, which being prolonged pas∣seth through the Center, is the least of those Lines which are drawn to the con¦vex Circumference LFIK. For exam¦ple, it is less than AI. Draw IB.

Demonstration. The Sides AI, IB, are greater than AB, (by the 20^th. of the 1^st.) wherefore taking away the equal Lines BI, BF, AF shall be less than AI.

4. AI is less than AK. Draw the Line BK

Demonstration. In the Triangle•• AIB, AKB, the sides AK, KB, ar•• greater than the Sides AI, IB, (by th•• 21^th. of the 1^st.) wherefore taking away the equal sides, BK, BI, there remain•• AI less than AK.

5. There can be drawn but two equal Lines; make the Angles ABL, ABK, equal.

Demonstration. The Triangles ABL, ABK, shall have their Bases AL, AK, equal (by the 4^th. of the 1^st.) but there cannot be drawn any other, which will not be either nearer or farther from AF, and which will not be either great∣er or lesser than AK, AL.