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any given finite straight line, that exceeds the half of the straight line joining the two given points.

For, let A, B be the two given points; and join A, B; and let CD be drawn bisecting AB at right; from Á, as a centre, at a distance equal to the given finite straight line, describe a circle, and let it cut CD in D; .. (Cor. 2.) D is equidistant from A and B; and .. a circle described from D, as a centre, at the distance DA, which (constr. E. 15. def. 1.) is equal to the given semidiameter, will pass through B.

PROP. IV.

8. THEOREM. If the three sides of a given triangle be bisected, the perpendiculars drawn to the sides, from the three several bisections, shall all meet in the same point: And that point is equidistant from the three angular points of the given triangle.

Let ABC be a given A, of which the three sides

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AB, AC, and CB are bisected in the points D, E

and F, respectively: The perpendiculars drawn to the several sides from D, E, F, shall all meet in a point that is equidistant from A, B and C.

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For, draw (E. 11. 1.) DG to AB, and EG 1 to AC, and let them meet in G: Join G, F. Then, (constr. and S. 3. 1. Cor. 2...) BG

and AG=CG; ... CG=BG.

AG,

Again, since (hyp.) BFCF (constr.) and F G

is common to the two

BFG, CFG, and that

BGCG,.. the BFG

CFG (E. 8. 1.); i.e.

(E. 10. def. 1.) GF is 1 to BC: And there cannot (E. 10. def. 1.) be drawn from F more than one straight line to BC. It is plain, therefore, that the perpendiculars drawn to the sides, from D, E and F, all meet in the same point G: And, since it has been shown that AG=BG=CG, the point G is equidistant from A, B and C.

PROP. V.

9. PROBLEM. To find a point, in a given plane, which shall be equidistant from three given points in the plane, that are not all in the same straight line.

Let A, B, C. be three given points, not all of them in the same straight line: It is required to find a point, that shall be equidistant from A, B and C.

Join A, B, and B, C, and C, A; bisect (E. 10. 1.) AB in D, and AC in E; draw (E. 11. 1.) from D

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and E, DG to AB, and EG 1 to AC, and let them meet in G.

Then, (S. 3. 1. Cor. 2.) the point G is equidistant from A, B, and C.

may

10. COR. By the help of this problem a circle be described about a given triangle; or so as that its circumference shall pass through any three given points that are not in the same straight line.

11. THEOREM.

PROP. VI.

There cannot be drawn more than two equal straight lines, to another straight line, from a given point without it.

Let A be a given point, without the given straight line BC: There cannot be drawn more than two equal straight lines, from A to BC. For, if it be possible, let AB-AG=AC; ..

(E. 5. 1.)

ABC;..

ACB

AGC

AGC: Also ACB=4
ABC; i. e. the exterior is

equal to the interior opposite 2, when the side

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BG, of the ▲ AGB, is produced: which (E. 16. 1.) is absurd.

12. COR. A circle cannot cut a straight line in more points than two.

PROP. VII.

13. THEOREM. The perpendicular let fall from the obtuse angle of an obtuse-angled triangle, or from any angle of an acute-angled triangle, upon the opposite side, falls within that side: But the perpendicular drawn to either of the sides containing the obtuse angle of an obtuse-angled triangle, from the angle opposite, falls without that side.

Let ABC be an obtuse-angled A, obtuse-angled at B, and let ABD be an acute-angled A: The perpendicular drawn from B to AC falls within AC; the perpendicular drawn from any other ▲ A, of the A ABC, to the opposite side BC, falls without BC; and the perpendicular drawn from

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any A, of the A ABD, to the opposite side

BD, falls within BD.

For, first, if it (E. 12. 1.) from A

be possible, let AG, drawn to BD, meet DB, produced, in G: Then, since (hyp.) the ABD is acute, the ABD is (E. 13. 1.) obtuse; and (constr.) the AGB is a right angle: Wherefore the two ABG, AGB of the A ABG are not less than two right angles; which (E. 17. 1.) is absurd. Therefore, the perpendicular drawn from A on BD cannot fall without BD. ner, it may be shewn, that the perpendicular drawn from B on the opposite side AC, of the obtuse-angled ▲ ABC, cannot fall without AC, and also that the perpendicular drawn from A, on the opposite side BC, of that A, cannot fall within BC.

And, in the same man

PROP. VIII.

14. THEOREM. If a straight line, meeting two other straight lines, makes the two interior angles

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