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PROPOSITION X.

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

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PROPOSITION XI.
Problem. To draw
a straight line at right
angles to a given
straight line from a
given point in the same. A D

CE
CoR. ( Argument ad absurdum).
Two right lines cannot have a common segment.

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PROPOSITION XII. Problem. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

PROPOSITION XIII. Theorem. The angles which one straight line makes with another on one side of it, are either two right angles, or are together equal to two right angles. Fig. 1.

Fig. 2.

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PROPOSITION XIV.

( Argument ad absurdum). Theorem. If, at a point in a straight line, two other straight lines on opposite sides of it make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

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Steps of the Demonstration. Suppose that bd is not in the same st. line with bc, but that BE is in the same st. line with it.

Then prove, on that supposition, 1. that ZS ABE + ABC = 2 rt. Zs, 2. that ZS ABE + ABC = Zs ABC + ABD, 3. that Z ABE = ABD, i. e. less = greater, which

shows the supposition to be false, 4. that similarly* no other line but bp is in the

same right line with Bc.

PROPOSITION XV. Theorem. If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

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Steps of the Demonstration. 1. Prove that ZS AEC + AED = 2 right ZS,

that Zs AED DEB = ZS AEC + AED, that L AEC

= L BEC, that similarly 2 AED = Z BEC.

* In every case in which we have employed the word “similarly," it has reference to the complete demonstration of the steps as given in Euclid.

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Steps of the Demonstration. 1. Prove that (in AS EAB, ECF) and ZBAC = LACF,

base AB = base FC, that l ACD > Z BAC, that similarly L BCG or ACD > LABC.

3.

PROPOSITION XVII.

Theorem. Any two angles of a triangle are together less than two right angles.

Steps of the Demonstration. 1. Prove that Zs DCA + ACB > ZS ACB + CBA,

that Zs ACB + CBA < 2 right Zs, 3. that similarly each of the other pairs of

Zs is < 2 right Zs.

3.

PROPOSITION XVIII. Theorem. The greater side of every triangle is opposite to the greater angle.

Steps of the Demonstration. 1. Prove that 2 ABD = L ADB,

that 2 ABD > L ACB, 3. that .. much more LABC > L ACB.

PROPOSITION XIX.

( Argument ad absurdum). Theorem. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

B Steps of the Demonstration. 1. Prove that ac cannot be = AB,

that ac cannot be < AB,
that :. Ac must be > AB.

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PROPOSITION XX.

Theorem. Any two sides of a triangle are together greater than the third.

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Steps of the Demonstration. 1. Prove that 2 ADC = L ACD,

that Z BCD > LACD,
that :: DB > BC,
that :. BA + AC > BC,
that similarly each of the other pairs of

sides is > the remaining side.

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