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(Prop. XIII.); .. the 3 ABC + ACB are < two rt. s.

3. In like manner it may be proved that the <S BAC+ ACB, and the S CAB + ABC, are <two rt. 4s.

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 4 at the base of an isosc. A are acute; and that, if one of a ▲ be right or obtuse, the two others must be acute.

PROP. XVIII. THEOR.

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

PART. ENUN.-Let ABC be any

the side AC is > than AB; then ABC shall be > the - АСВ.

CONST.-Because AC is > B

A

of which

AB, cut off AD = AB. (Prop. III.) Join BD.
DEMONST.-1. Then the exterior

of the DBC is

XVI.)

ADB

interior DCB. (Prop.

2. Now, because AB = AD, .'. ≤ ABD = 2

ADB (Prop. V.); .. ≤ ABD is >

3. ., a fortiori, ABC is > the

ACB.

ACB.

Wherefore the greater side &c.—Q. E. D.

It appears from this Proposition that all the 4s of a scalene▲ are unequal. If BC be also greater than AB, it

may be proved, in like manner, that the ACB; and that as BC is > or <

than the

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BAC is greater

AC, the ▲ BAC

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 ▲, of which

the ABC is > the

ACB;

then the side AC shall be >

AB.

DEMONST.-For, if not >, AC must either be = AB, or

B

1. It is not, for then the be ACB (Prop. V.), but it 2. Neither is it <, for then the be the ACB (Prop. XVIII.), 3... the side AC is neither

i. e. AC is > AB.

A

C

than it.
ABC would

is not.

ABC would but it is not. nor < 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.

C

PART. ENUN. and CONST.-Let AB be a st. line of unlimited length, c the gn. pt. From c draw CD 1 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

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B

E F G

... CED is < a rt. 4 (Prop. XIII.), < ▲ CDE; .'. CD is <CE. (Prop. XIX.)

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

3. Again, because the CEF is an obtuse ; .. CFE is an acute ; .. 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.

the 4

PROP. N. THEOR.

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

PART. ENUN.-Let ABC be a A. is greater than the difference between the other two sides.

CONST.-Let AC be > AB, and make AD = AB. (Prop. III.)

BD (Post. 1).

Then any of its sides

Join

B

D

C

DEMONST.-1. Because AB AD, .. the ABD = L

ADB. (Prop. V.)

2. Now the exterior / BDC of ▲ ABD is the interior Z ABD (Prop. XVI.), i. e. > ▲ ADB; .., a fortiori, > Z DBC (Prop. XVI.); .. the side BC is the side DC (Prop. XIX.), > AC

AD, AC - AB.

3. In the same way it may be shown that AB is > AC — BC, and AC > BC

- AB.

Wherefore any side &c.-Q. E. D.

PROP. XX. THEOR.

GEN. ENUN.-Any two sides of a triangle are together greater than the third side.

BC, AB + BC > AC,

PART. ENUN.-Let ABC be any Δ. Then the sum of any two of its sides is the third side, viz., BA + AC is and BCCA > AB. CONST.-Produce BA to the pt. D (Post. 2), making AD AC. (Prop. III.) Join DC. Then, DEMONST.-1. Be

cause DA = ac, :. 2

B

ADC = ACD (Prop. V.); but the > the ACD (Ax. 9); .. the

BCD is

BCD is >

the ADC; and the side BD is > the side BC. (Prop. XIX.)

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

3. In the same manner, it may be proved that ABBC is > AC, and that BC + CA is > AB. Wherefore, any two sides &c.-Q. E. D.

This

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. 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 + BC are > AC, produce CB to D, make BD

= AB, and join AD. Then be

[blocks in formation]

A

the DAC is > CDA; .. in the ▲ ADC, the side DC is side AC (Prop. XIX.) Now DC = DB + BC = AB + BC (Const.); . ;.. AB + BC is > AC.

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BC.

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

A

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 ▲ ADB, the side AB is >

BD.

B

C

D

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 ▲, 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 Bс 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).

B

C

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 A by joining AC, AD, AE, and so on for any number

F

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