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SECTION IV. PROPOSITIONS XVIII.-XXV.
Inequalities of Parts of Triangles, and Accessory
55. EXERCISE XXIII. In the annexed diagram, PQR is an equilateral
P triangle ; it is referred to in 1–7.
1. Name an exterior angle of the triangle PQS, and also of the triangle QRS.
is 2. Name the angle which is greater than angle P, by I. 16.
3. Also that which is greater than angle R. Q 4. Find two angles in the figure which are less than R, and prove them so.
5. From Questions 3 and 4 deduce which is the greater angle, PSQ or PQS.
6. Show that in magnitude angle P is intermediate between angles SQR and QSR.
7. Show that angle P is less than either angle at S.
8. Given a triangle PQS having the side PS less than the side PQ, show that the angle opposite PS is less than the angle opposite PQ.
(For construction, produce PS to R, making PR equal to PQ, and join QR.)
9. In the figure to Prop. XVIII., given AD equal to AB, prove that the angle ADB is intermediate in magnitude between the angles ABC and C.
10. If, in the same figure, the angle C is greater than DBC, can we say that DB and DC are equal? If not, give the reason fully, with reference.
56. PROPOSITION XVIII.-THEOREM. If a triangle have two sides unequal, the angle which is opposite
the greater side is greater than the angle opposite the less.
Let ABC be a triangle, of which the side AC is greater than the side AB.
Then the angle ABC shall be greater than the angle ACB.
From AC the greater cut off AD equal to AB the less, (I. 3) and join BD.
Then, because AD is equal to AB in the triangle ABD, the angle ABD is equal to the angle ADB;
(I. 5) and ADB is an exterior angle of the triangle BCD; therefore it is greater than the interior opposite angle DCB;
(I. 16) therefore, also, the angle ABD is greater than DCB; even more, then, is the angle ABC greater than the angle ACB. Therefore, if a triangle have, &c.
Q. E. D.
57. PROPOSITION XIX.-THEOREM. If a triangle have two angles unequal, the side which is opposite the greater angle is greater than the side which is opposite the less.
Let ABC be a triangle, of which the angle ABC is greater than the angle ACB,
Then the side AC shall be greater than the side AB.
For, if AC be not greater than AB,
If AC were equal to AB, then the angles ABC and ACB would be equal; (I. 5) but they are not;
(Нур.) therefore AC is not equal to AB.
Again, if AC were less than AB, then the angle ABC would be less than ACB; (I. 18) but it is not;
(Hyp.) therefore AC is not less than AB.
Hence, AC is neither equal to nor less than AB; therefore AC is greater than AB. Wherefore, if a triangle have, &c.
Q. E. D.
EXERCISE XXIV. 1. In a scalene triangle, prove that each pair of angles is unequal.
2. In a right-angled triangle, prove that the hypotenuse is the greatest side.
3. If two straight lines be drawn from a point to meet a straight line, one of which is perpendicular to it, show that this perpendicular must subtend an acute angle where the oblique meets the straight line to which it is drawn.
4. Show also that the perpendicular must be less than the oblique.
5. If there were two oblique lines on the same side of the perpendicular, show that the one nearer to the perpendicular is the lesser of the two.
6. In the annexed figure, AD is perpendicular to BC; prove that CD is less than AC, and BD less than AB.
7. Show from these inequalities that BC is less than BA and AC together.
8. In the next figure, AD bisects the angle BAC. Show that the angle CDA is greater than the angle CAD. (Apply I. 16.)
9. Draw an inference with regard to the comparative magnitudes of CD, CA.
10. Prove that BD is less than BA.
11. From Questions 9 and 10 deduce a method of proving that BC must be less than BA and B AC together.
12. By bisecting the angle C, show that BC and CA are together greater than AB.
58. PROPOSITION XX.—THEOREM. Any two sides of a triangle are together greater than the third side.
Let ABC be any triangle.
Any two of its sides shall be together greater than the third ; namely, BA, AC greater than BC; AB, BC greater than CA; and BC, CA greater than AB.
Produce BA to D, making AD equal to AC, and join DC.
(I. 3) Then, because AD is equal to AC, the angle ACD is equal to the angle ADC ;
(1. 5) but the angle BCD is greater than ACD; therefore, also, BCD is greater than ADC; that is, in the triangle BCD,
the angle BCD is greater than the angle BDC. But the greater angle is subtended by the greater side; (I. 19) therefore the side BD is greater than the side BC. But BD is equal to BA and AC;
therefore BA and AC are together greater than BC.
In the same manner, it may be demonstrated that the sides AB, BC are greater than CA, and also that BC, CA are greater than AB. Wherefore any two sides, &c.
Q. E. D. 59. PROPOSITION XXI.—THEOREM. If from the ends of a side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the two sides of the triangle, but shall contain a greater angle.
Let ABC be a triangle, and D a point within it, and from B, C, the ends of the side BC, let two straight lines BD, CD be drawn to the point D.
Then BD and DC shall be less than BA and AC, but shall contain an angle BDC greater than the angle BAC.
Produce BD to meet the side AO in E.
Since two sides of a triangle are greater than the third side,
(I. 20) therefore, in the triangle ABE, the two sides BA, AE are greater than BE; to each of these unequals add EC ;
then the sides BA, AC are greater than BE, EC. (Ax. 4)
Again, in the triangle CED, the two sides DE, EC are greater than DC; (I. 20) add BD to each of these unequals;
then the sides BE, EC are greater than BD, DC. (Ax. 4) But it has been shown that BA, AC are greater than BE, EC; even more, then, are BA, AC greater than BD, DC.
Again, because BDC is an exterior angle of the triangle
CDE, it is greater than the interior opposite angle DEC. (I. 16) And, because DEC is an exterior angle of the triangle BAE,
it is greater than the interior opposite angle BAC.
But it has been shown, that the angle BDO is greater than the angle DEC; even more, then, is the angle BDC greater than BAC. Therefore, if from the ends, &c.
Q. E. D. EXERCISE XXV. 1. Give the proof of XX. that BA and AC are greater than BC, by producing CA instead of BA.
2. Give the proof in full that BC, CA are greater than BA.
3. Twice the longest side of a triangle must be less than the sum of the three sides.
4. Prove Prop. XXI. by producing CD to meet AB in E in the con. struction.
5. Prove that which relates to vertical angles, by means of joining AD, and then producing AD.
6. In an isosceles triangle, show that the base cannot be twice as great as one of the equal sides.
7. Given three straight lines respectively equal to the sides of a triangle; if the first be half of the second, show that the third must be greater than the first.
8. Given three pieces of straight wire, which are to be placed upon a table in the form of a triangle ; if this cannot be done, what will be the cause ?