« ΠροηγούμενηΣυνέχεια »
1. Two straight lines AB, CD intersect at E; show that the bisectors of the angles AED, BEC are in the same straight line.
2. From two given points on the same side of a given line, draw two lines which shall meet in that line and make equal angles with it.
PROPOSITION XVI. THEOREM.
If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.
Let ABC be A, and let any side BC be produced to D.
Then the ext. _ ACD shall be greater than either of the int. opp. Es CBA or BAC.
G Constr. Bisect AC at E (i. 10). Join BE, and produce
BE to F, making EF=BE, (i. 3) and join FC. Dem. :: AE= EC, and BE = EF; (constr.) :, the two
sides AE, EB= the two CE, EF, each to each in As ABE, CFE; and L AEB = L CEF; (i. 15) .:. Δ ΑΕΒ =ΔCEF;
(i. 4) and the rem. Zs of the one = the rem. Zs of the other, each to each, to which the = sides are opposite;
... BAE= ECF; but _ ECD or ACD is greater than _ ECF; (ax. 9)
.: L ACD is greater than _ BAE or BAC. In the same manner, if BC be bisected, and AC produced to G; it may be shown that L BCG, that is L ACD (i. 15) is greater than 2 ABC. Therefore, if one side of a triangle, &c.
Q. E. D. [In repeating or writing out this prop., do not omit to show how Z ACD may be proved greater than 2 ABC-a common mistake.]
Any two exterior angles of a triangle are together greater than two right angles.
PROPOSITION XVII. THEOREM.
Any two angles of a triangle are together less than two right angles.
Let ABC be A.
Then any two of its s together shall be less than two rt. Ls.
Constr. Produce any side BC to D. Dem. Then, :: ACD is ext. L of A ABC, it is greater than int. opp. L ABC;
(i. 16) to each of these unequals add L ACB; :: ZS ACD, ACB are greater than s ABC, ACB;
(ax. 4) but ZS ACD, ACB together = two rt. Zs; (i. 13) .. Ls ABC, ACB together are less that two rt. Zs.
In like manner it may be demonstrated that ZS BAC, ACB are together less than two rt. Zs, as also Zs CAB, ABC.
Therefore, any two angles of a triangle, &c. Q. E. D.
[This theorem proceeds at once as a corollary to prop. 16. The learner should make the construction and work out the prop. for the other angles.]
PROPOSITION XVIII. THEOREM.
The greater side of every triangle is opposite to the greater angle.
[That is, the greater side of every triangle has the greater angle opposite to it.]
Let ABC be A, of which AC is greater than AB.
Constr. Since AC is greater than AB, make AD= AB,
(i. 3) and join BD. Dem. Then, ::: AD= AB .. _ ADB = L ABD; (i. 5)
but the side CD of A DBC is produced to A,
(i. 16) but L ADB has been proved = L ABD,
.: L. ABD is greater than _ DCB; much more is 2 ABC greater than _ ACB. Therefore, the greater side, &c.
Q. E. D.
PROPOSITION XIX. THEOREM.
The greater angle of every triangle is subtended by the greater side, or, has the greater side opposite to it.
Let ABC be 8 of which _ ABC is greater than _ BCA.
Then AC is greater than AB.
Dem. For if AC be not greater than AB, AC must either be equal to or less than AB.
If AC= AB, then _ ABC= L ACB; (i. 5) but it is not equal, (hyp.) :. AC is not = AB. Again, if AC were less than AB, then _ ABC would be less than L ACB;
(i. 18) but it is not less, :, AC is not less than AB; and AC has been shown to be not = AB,
: . AC is greater than AB. Therefore, the greater angle, &c.
Q. E. D. [Beginners are apt to confuse together props. 18 and 19: at the examination of one of our foremost public schools this year, 14 boys out of 40 in the first class wrote out the wrong prop. There is in fact a little ambiguity, and care should be taken to observe the distinction. The first, in which the hypothesis is that one side is greater than one side, is demonstrated directly; the second, when it is given that one angle in the one is greater than one angle in the other, is demonstrated indirectly; they thus follow, in this respect, the order of props. 5 and 6.)