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If from a point without a straight line a perpendicular is

let fall on the line, and oblique lines are drawn to dif

ferent points of it: The perpendicular is shorter than any oblique line. 2o. Any tuo oblique lines that meet the given line at points

equally distant from the foot of the perpendicular, are

equal. Of tuo oblique lines that meet the given line at points

unequally distant from the foot of the perpendicular, the one which meets it at the greater distance is the longer.

a

a

B

Let A be given point,

point, DE given straight line, AB a perpendicular to DE, and AD, AC, AE oblique lines,

D

E BC being equal to BE, and BD greater than BC. Then AB is less than any of the oblique lines, AC is equal to AE, and AD greater than AC.

Prolong AB until BF is equal to AB, and draw FC, FD.

F

1o. In the triangles ABC, FBC, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and FBC equal, because both are right angles: hence, FC is equal to AC (P. V.). But, AF is shorter than ACF (A. 12): hence, AB, the half of AF, is shorter than AC, the half of ACF; which was to be proved.

2°. In the triangles ABC and ABE, we have the side BC equal to BE, by hypothesis, the side AB common, and the included angles ABC and ABE equal, because both are right angles: hence, AC is equal to AE; which was to be proved.

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3°. It may be shown, as in the first case, that AD is equal to DF. Then, because the point clies within the triangle ADF, the

of the lines AD and DF is greater than the sum of the lines AC and CF (P. VIII.): hence, AD, the half of ADF, is greater than AC, the half of ACF; which was to be proved.

Cor. 1. The perpendicular is the shortest distance from a point to a line.

Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible.

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F

If a perpendicular is drawn to a given straight line at its

middle point: . Any point of the perpendicular is equally distant from

the extremities of the line: 2o. Any point, without the perpendicular, is unequally dis

tant from the extremities.

· Let AB be a given straight line, C its middle point, and EF the perpendicular. Then any point of EF is equally distant from A and B; and any point without EF, is unequally distant from A and B.

1°. From any point of EF, as D, draw the lines DA and DB. Then DA and DB

E are equal (P. XV.): hence, D is equally distant from A and B; which was to be proved.

DI

A

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2. From any point without EF, as I draw IA and

, IB. One of these lines, as IA will cut EF in

, point D; draw DB. Then, from what has just been shown, DA and DB are

F equal; but 1B is less than the sum of ID and DB (P. VII.); and because the sum of ID and DB is equal to the sum

of ID and DA, or TA, we have IB less than IA: hence, 1 is unequally distant from A and B; which was to be proved.

Cor. If a straight line, EF, has two of its points, E and F, each equally distant from A and B, it is perpen

B dicular to the line AB at its middle point.

A

B

E

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A

D

If two right-angled triangles have the hypothenuse and a

side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal in all respects.

Let the right-angled triangles ABC and DEF have the hypothenuse AC equal to DF, and the side AB equal to DE: then the triangles are equal in all respects.

If the side BC is equal to EF, the triangles are equal, in accordance with Proposition X. Let us suppose then, that BC and EF are unequal, and that BC is the longer. On BC lay off BG equal to EF, and draw AG. The triangles ABG and DEF have AB equal to DE, by hypothesis, BG equal to EF, by construction, and the angles B and E

B

equal, because both are right angles; consequently, AG is equal to DF (P. V.). But, AC is equal to DF, by hypothesis: hence, AG and AC are equal, which is impossible (P. XV.). The hypothesis that BC and EF are unequal, is, therefore, absurd: hence, the triangles have all their sides equal, each to each, and are, consequently, equal in all respects; which was to be proved.

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If tuo straight lines are perpendicular to a third straight

line, they are parallel.

B

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Let the two lines AC, BD, be perpendicular to AB: then they are parallel.

For, if they could meet in a point O, there would be two perpendiculars OA, OB, drawn from the same point

-С to the same straight line; which is impossible (P. XIV.): hence, the lines are parallel ; which was to be proved.

A

DEFINITIONS.

E

If a straight line EF intersect two other straight lines AB and CD, it is called a secant,

, with respect to them. The eight angles formed about the points of intersection have

different names, with respect to each other.

B

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D

1°. INTERIOR ANGLES ON THE SAME

F SIDE, are those that lie on the same side of the secant and within the other two lines. Thus, BGH and GHD are interior angles on the same side.

2°. EXTERIOR ANGLES ON THE SAME SIDE are those that lie on the same side of the secant and without the other two lines. Thus, EGB

EGB and DHF are exterior angles on the same

E -side.

A

В 3°, ALTERNATE ANGLES

are those that lie on opposite sides of the

-D

H secant and within the other two

F lines, but not adjacent. Thus, AGH and GHD are alternate angles.

4°. ALTERNATE EXTERIOR ANGLES are those that lie on opposite sides of the secant and without the other two lines. Thus, AGE and FHD are alternate exterior angles.

5o. OPPOSITE EXTERIOR AND INTERIOR ANGLES

are those that lie on the same side of the secant, the one within and the other without the other two lines, but not adjacent. Thus, EGB and GHD are opposite exterior and interior angles.

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If tuo straight lines meet a third straight line, making

the sum of the interior (mgles on the same sile equal to tuo right angles, the two lines are parallel.

Let the lines KC and HD meet the line BA, making the sum of the angles BAC and ABD equal to two right angles; then KC and HD are parallel.

Through G, the middle point of AB, draw

draw GF
GF perpendicular

E B

Hto KC, and prolong it to E.

The sum of the angles GBE and GBD is equal to two right

G

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-C

A F

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