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If, in the triangle ABC, the angle CAB be greater than ACB; its opposite side BC is greater than AB.

For if BC be not greater than AB, it must be either equal or less. But it cannot be equal, be- T;

cause the angle CAB would then be equal to ACB (I. 10.); nor can BC be less than AB, for then AB would be greater than

BC, and consequently (I. 12.) the angle * C ACB would be greater than CAB, or CAB less than ACB, which is absurd. The side BC being thus neither equal to AB, nor less than it, must therefore be greater than AB.


Two sides of a triangle are together greater than the third side, - - - -

The two sides AB and BC of the triangle ABC are together greater than the third side AC. For produce AB until DB be equal to the side BC, and join CD. Because BC is equal to BD, the angle BCD is equal to BDC (I. 10.); but the angle ACD -

is greater than BCD, and therefore I) greater than BDC, or ADC; Con- B sequently the opposite side AD is greater than AC (I, 13.); and since 2^ AD is equal to AB and BD, or to K C

AB and BC, the two sides AB and
BC are together greater than the third side AC.


The difference between two sides of a triangle is less than the third side.

Let the side AC be greater than AB, and from it cut off a part AE equal to AB; the remainder EC is less than the third side BC. B

For the two sides AB and BC are together greater than AC (I. 14.); take away the equal lines AB and AE, and there remains BC greater than EC, or EC is less than BC.

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Two straight lines drawn to a point within a triangle from the extremities of its base, are together less than the sides of the triangle, but contain a greater angle. -

The straight lines AD and CD, projected to a point D within the triangle ABC from the extremities of the base AC, are together less than the sides AB and CB of the triangle, but contain a greater angle.

For produce AD to meet CB in E. The two sides AB and BE of the triangle ABE are greater w than the third side AE (I. 14.); add EC to each, and AB, BE, EC, or AB and TE BC, are greater than AE and EC. But the sides CE and ED of the triangle DEC are (I, 14.) greater than DC, and

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consequently CE, ED, together with DA, or CE and EA, are greater than CD and DA. Wherefore the sides AB and BC, being greater than AE and EC, which are themselves greater than AD and DC, must be still greater than AD and DC, or the lines AD and DC are less than AB and BC, the sides of the triangle. Again, the angle ADC, being the exterior angle of the triangle DCE, is greater than DEC (I. 8.); and, for the same reason, DEC is greater than ABE, the opposite interior angle of the triangle EAB. Consequently ADC is still greater than ABE or ABC.


If straight lines be drawn from the same point to another straight line, the perpendicular is the shortest of them all; the lines equidistant from it on both sides are equal; and those more remote are greater than such as are nearer.

Of the straight lines CG, CE, CD, and CF drawn from a given point C to the straight line AB, the perpendicular CD is the least, the equidistant lines CE and CF are equal, but the remoter line CG is greater than either of these tWO.

For the right angle CDE, equal to CDF, is (I. 8.) greater than the interior angle CFD of the triangle DCF, and consequently the opposite

side CF is (I, 13.) greater
than CD, or CD is less than -
CF. - 2. 2 A-

But if ED be equal to FD, A GTIETID F B CD being common to the two triangles ECD and FCD, and the contained angles CDE and CDF equal; these triangles (I. 3.) are equal, and consequently their bases CE and CF are equal. Again, because GCD is a right-angled triangle, the angle CGD or CGE is (I. 9. cor.) acute, and, for the same reason, the angle CED of the triangle CDE is acute, and consequently its adjacent angle CEG is obtuse. Wherefore CEG, being greater than a right angle, is still greater than CGE, and the opposite side CG must be greater (I. 13.) than CE. - Cor. Hence only a single perpendicular CD can be let fall from the same point C upon a given straight line AB; and hence also a pair only of equal straight lines greater than CD can at once be extended from C to AB, making on the same side, the one an obtuse angle CEA, and the other an acute angle CFA.—As the term distance signifies the shortest road, the distance between two points, is the straight line which joins them; and the distance from a point to a straight line, is the perpendicular let fall upon it.


If two sides of one triangle be respectively equal

to those of another, but contain a greater angle;

the base also of the former will be greater than that of the latter. -

In the triangles ABC and DEF, let the sides AB and BC be equal to DE and EF, but the angle ABC greater than DEF; then is the base AC greater than DF,


For, suppose AB one of the sides to be not greater than BC or EF, and (I. 4.) draw BG equal to EF making an angle ABG equal to DEF, join AG and GC,

Because AB and BG are equal to DE and EF, and the contained angle ABG is equal to DEF; the triangles ABG and DEF (I. 3.) are equal, and have equal bases AG and DF. -

First, let the triangles ABC and DEF be isosceles. Since the side AB is equal

to BC, the angle BAC (I.10.) B To. is equal to BCA ; but (I. 8.) the angle BHC is greater AS II \ D than the interior angle BAH o G. C. IE

or BCH, and consequently (I. 13.) the side BC or BG is greater than BH, or the point G lies beyond H.

Next, suppose the side

BC or EF to be greater 3B R

than AB or DE. Where

fore (I.12.) the angle BAC A D

is greater than BCA ; but

(I. 8.) the exterior angle Gr C 16

BHC of the triangle ABH is greater than BAH or BAC, and hence still greater than BCA or BCH ; consequently the side BC or EF is (I, 13.) greater than BH.

In every case, therefore, the point G must lie below the base AC. But the triangle GBC being evidently isosceles, its angles BGC and BCG (I. 10.) are equal. Whence the angle AGC, being greater than BGC or BCG, which a- gain is greater than ACG, must be still greater than ACG; and therefore the opposite side AC is (I, 13.) greater than AG or DF.

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