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PROPOSITIONS 1-26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS.

PROPOSITION XI.

To draw a straight line at right angles to a given straight line, from a given point in the same. be the given straight line, and a given point in it.

Let

It is required to draw a straight line from the point

at right angles to

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ides

ase

; therefore the angle

and join

is equal to
are equal to the two sides

Then

drawn from the point

shall be at right angles to

and

is common to the two triangles

; the two

each to each; and the base is equal to the and these two angles are adjacent

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is equal to the angle

ngles. But when the two adjacent angles which one straight line makes with another straight line, are qual to one another, each of them is called a right angle: therefore each of the angles

s a right angle.

Wherefore from the given point , in the given straight line

ngles to

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COR. By help of this problem, it may be demonstrated that two straight lines cannot have a ommon segment. If it be possible, let the segment

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is a straight line, therefore is equal to the angle

the less equal to the greater angle, which is

Therefore two straight lines cannot have a common segment.

PROPOSITIONS 1-26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS.

PROPOSITION X.

To bisect a given finite straight line, that is, to divide it into two equal parts.

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shall be cut into two equal parts in the point

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PROPOSITIONS 1-26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS.

PROPOSITION IX.

Let

To bisect a given rectilineal angle, that is, to divide it into two equal angles.

be the given rectilineal angle.

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