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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 ▲ ABC, to the opposite side BC, falls without BC; and the perpendicular drawn from

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any A, of the ▲ ABD, to the opposite side BD, falls within BD.

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For, first, if it be possible, let AG, drawn (E. 12. 1.) from A 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. And, in the same manner, it may be shewn, that the perpendicular drawn from B on the opposite side AC, of the obtuse-angled A 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.

PROP. VIII.

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

on the same side of it not less than two right angles, these lines shall never meet on that side, if produced ever so far.

For, if it be possible, let two straight lines meet, which make, with another straight line, the two interior angles, on the same side, not less than two right: Then it is plain, that the three straight lines will thus include a A, two of which are not less than two right angles; which (E. 17. 1.) is absurd. Wherefore, the two straight lines cannot meet, on that side of the straight line, on which they make the two interior

than two right.

not less

15. COR. Two straight lines, which are both perpendicular to the same straight line, are parallel to each other.

PROP. IX.

16. THEOREM. The three sides of a triangle taken together, exceed the double of any one side, and are less than the double of any two sides.

For, since (E. 20. 1.) any two sides of a ▲ are > the third, if the third side be added both to those two and to itself; it is evident that the three sides are, together, > the double of the third.

Again, since (E. 20. 1.) any side of a ▲ is < the other two, if the other two be added both to that side, and to themselves, it is evident, that the

three sides are, together, < than the double of the other two.

PROP. X.

17. THEOREM. Any side of a triangle is greater than the difference between the other two sides.

If, the A be equilateral, or isosceles, the proposition is manifestly true. But let it be a scalene. A: Then, since (E. 20. 1.) any two sides of the A are the third, if either of those two be taken from that third side, it is plain that the remaining side is greater than the difference of the other

two.

PROP. XI.

18. THEOREM. Any one side of a rectilineal figure is less than the aggregate of the remaining sides.

Let ABCD be a given rectilineal figure: Any

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B

one side, as BC, is less than the aggregate of the remaining sides.

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For, first, let the figure be quadrilateral; and join B,D: Then (E. 20. 1.) BD + DC > BC; and, BA+AD > BD; .;. BA+AD+DC>BD+DC; much more, then, is BA + AD + DC > BC.

And the proposition may, in the same manner, be proved to be true, when the figure has more than four sides.

PROP. XII.

19. THEOREM. The two sides of a triangle are together, greater than the double of the straight line which joins the vertex and the bisection of the base.

Let ABC be any given ▲, and let AD be the

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straight line joining the vertex A, and the bisection, D, of the base BC: AB+ AC > 2AD. Produce. AD to E, and cut off (E. 3. 1.) DE= AD; also, join B, E.

Then since (hyp.) BD=DC, and (constr.) AD=DE, the two sides BD, DE, of the ▲ BDE, are equal to the two sides AD, DC of the A ADC;

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