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(Prop. XIII.); ·:. the ZS ABC + ACB are < two rt. Zs.

3. In like manner it may be proved that the <$ BAC + ACB, and the ZR CAB + ABC, are < two rt. Zs.

Therefore any two angles &c.-Q. E. D. This and the preceding Proposition are carried further in Prop. XXXII. It will be well to prove it in all its cases by producing each side successively.

The following are Corollaries:--That the 28 at the base of an isosc. A are acute; and that, if one / of a Abe right or obtuse, the two others must be acute.

PROP. XVIII. THEOR.

A

GEN. ENUN.—The greater side of every triangle is opposite to the greater angle.

PART. ENUN.—Let ABC be any a of which the side ac is > than AB; then < ABC shall be > the Z ACB.

Const.–Because AC is > B
AB, cut off AD = AB. (Prop. III.) Join BD.

DEMONST. -1. Then the exterior Z ADB of the a DBC is > interior _ DCB. (Prop. XVI.)

2. Now, because AB = AD, .. 2 ABD = 2 ADB (Prop. V.); .. 2 ABD is > < ACB.

3. .., a fortiori, 2 ABC is > the 2 ACB. Wherefore the greater side &c.—Q. E. D. It appears from this Proposition that all the 4s of a scalene A are unequal. If BC be also greater than AB, it

may be proved, in like manner, that the / Bac is greater than the L ACB; and that as Bc is > or < AC, the BAC is > or < the ABC.

PROP. XIX. THEOR.

GEN. ENUN.—The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

PART. ENUN.-Let ABC be a A, of which the < ABC is > the < ACB; then the side ac shall be >

А

AB.

B

be

DEMONST.–For, if not > AC must either be

= AB, or < than it. 1. It is not =, for then the < ABC would

= < ACB (Prop. V.), but it is not. 2. Neither is it <, for then the 2 ABC would be < the < ACB (Prop. XVIII.), but it is not.

3. ... the side ac is neither = nor < AB; i.e. Ac is > AB.

Wherefore the greater angle &c.—Q. E. D. In like manner, the relative lengths of the other sides may be proved; and hence may be deduced the following:

PROP. M. THEOR.

GEN. ENUN.-Of all straight lines which can be drawn from a given point to a straight line of unlimited length, a perpendicular is the shortest, and that which is nearer to the perpendicular is shorter than one more remote.

с

PART. Enun, and Const.-Let AB be a st. line of unlimited length, c the gn. pt. From c draw CD I to AB (Prop. XII.), and draw ce, CF, CG, &c. to any other pts. along the line AB; then cd is the shortest line which can be drawn from c to AB, CE shorter than CF, CF than cg, and so on. DEMONST.-1. Because L

A.

B CDE is a rt. L, the exterior,

F CEF is > a rt. 2 (Prop. XVI.); .:. cep is < a rt. 4 (Prop. XIII.), < LCDE; .. cd is < CE. (Prop. XIX.)

2. In the same manner it may be shown that cp is < every other line drawn from c to AB.

3. Again, because the / CEF is an obtuse L; .. the L CFE is an acute L; .. CE is less CF. (Prop. XIX.)

4. In the same manner, cf is < cG, &c., &c. Therefore, of all st. lines &c.—Q. E. D.

D

.E

G

PROP. N. THEOR.

A

D

B

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GEN. ENUN.—Any side of a triangle is greater than the difference between the other two sides.

Part. ENUN.- Let ABC be a A. Then any of its sides is greater than the difference between the other two sides.

CONST. — Let AC be > AB, and make AD = AB. (Prop. III.) Join BD (Post. 1).

DEMONST.-1. Because AB = AD, .. the 2 ABD = L ADB. (Prop. V.)

2. Now the exterior / BDC of A ABD is > the interior

ABD (Prop. XVI.), i. e. > LADB; . ., a fortiori, > | DBC (Prop. XVI.); .. the side bc is the side dc (Prop. XIX.), > AC AD, > AC

3. In the same way it may be shown that ab is > AC – BC, and AC > BC Wherefore any side &c.—Q. E. D.

AB,

AB.

sum

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B

PROP. XX. THEOR. Gen. Enun.—Any two sides of a triangle are together greater than the third side.

Part. Enun.—Let ABC be any A. Then the of

any two of its sides is > the third side, viz., BA + AC is BC, AB + BC AC, and BC + CA > AB.

Const.–Produce BA to the pt. D (Post. 2), making AD= Ac. (Prop. III.) Join pc. Then,

DEMONST. -1. Because DA = AC, .. Z ADC < ACD (Prop. V.); but the < BCD is > the < ACD (Ax. 9); .. the BCD is > the < ADC; and the side bp is > the side BC. (Prop. XIX.)

2. But BD = BA + AD = BA + AC (Const.); .. BA + AC is > than BC.

3. In the same manner, it may be proved that AB + Bc is > AC, and that BC + ca is > AB.

Wherefore, any two sides &c.-Q. E. D. This is one of those Propositions of which it has been said that the proof is unnecessary, since any person would readily admit its truth without any demonstration. This may be very true : but the proof is still essential to the completeness of a connected mathematical system ; and the student must be able to work the two remaining cases.

In order, therefore, to prove that AB AC, produce co to D, make bo = AB, and join AD. Then be

BD, .. LBDA | BAD (Prop. V.): but the 2 DAC is > the BAD (Ax. 9);..

+ BC are >

A

cause BA

D

с

+

A

the DAC is > LCDA; .. in the A ADC, the side dc is > side ac (Prop. XIX.) Now DC = DB + BC = AB + BC (Const.); .. AB + BC is > AC.

Again, to prove that BC CA are > AB, produce AC to D, make cd = BC, and join BD. Then, because cd = BC, .'. L CBD = CDB (Prop. V.): but the . ABD is > 1 CBD (Ax. 9); .. the 4 ABD is > LADB; .. AD is > AB (Prop. XIX.): but ad + CD = AC + CB; .. AC + CB is > AB.

Another solution of the Pro. position, without producing the side, may not be unacceptable.

B

с

= AC

D

Thus:-

BC.

B

с

D

Let ABC be a A. Any two sides, as BA + AC, are >

Bisect the / Bac by the st. line AD (Prop. IX.); then the exterior / ADB of the ^ ADC is > the interior opposite / DAC (Prop. XVI.); .. also the , ADB is > the [ DAB; .. in the A ADB, the side AB is > BD.

In like manner it may be shown that ac is > DC; .. AB + AC are > BD + DC, i.e. > BC.

Of this Proposition the following are Corollaries :

Cor. 1.-The three sides of a A, taken together, exceed the double of any one side, but are less than the double of any two sides. For the sides AB + AC are > BC; add bc to each ; .'. AB + AC + BC are > twice Bc. But the side AB is < AC + BC; add AC + BC to each; .. AB + AC + BC is < twice AC + twice BC, or 2 (AC + BC).

Cor. 2.-Any one side of a rectilineal figure is less than the sum of the remaining sides.

Let ABCDEF be any rectilineal figure; divide it into As by joining AC, AD, AE, and so on for any number

F

B

с

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