( Argument ad absurdum).' Theorem. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite upon the same side; and the two interior angles upon the same side together equal to two right angles.



Steps of the Demonstration. Suppose that _ AGH # GhD, and that Agh > GHD,

Then prove, on that supposition, 1. that _ AGH + BGH > ZS BGH + GHD, 2. that s BGH + GHD < 2 right ZS, 3. that :: AB, CD would meet if produced far

enough*, which shows the supposition to be false,

and .. that _ AGH = 2 GHD. 4. Prove that ZS EGB = LGHD, 5. that ZS EGB + BGH = US BGH + GHD, 6. that ZS BGH + GhD = 2 right 8.

* The proof of this proposition depends on what is called the 12th Axiom, which, however, is so far from being an Axiom, that it is a Theorem quite as much in need of demonstration as that which it is here brought forward to establish. The author is happy to state that the following proof of it has been pronounced to be generally satisfactory by most competent authority

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Proved by showing that the alternate ZS AGK, GKD are equal.





PROPOSITION. If a right line meet two right lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these right lines being continually produced, shall at length meet on that side on which are the angles which are less than two right angles.Let the right line Ef meet the two right lines AB, CD; and

let the ZS BKL, KLD be

together < 2 right Zs; M Go

then shall AB, CB meet, if produced far enough, towards the points B

and D. Since lines which, when they are produced ever so far, and do not meet, are parallel lines ; it is plain that lines which are not parallel will meet, if produced sufficiently; for, if not, they would be parallel by the definition.

Now, let gh, a line passing through the point « in which EF cuts AB be such that ZS HKL + Z KLD = 2 right Zs,

:. by Proposition XXVIII. GH || CD,



.. AB is not || CD.
For it cuts GH, which is || CD, in the point i,

* An axiom is here implied, which is as self-evident as any employed in Euclid ; viz., that through the same point there cannot be drawn two straight lines which shall both be parallel to the same straight line.




Problem. To draw a straight line through a given point parallel to a Ĉ given straight line.


PROPOSITION XXXII. Theorem. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.




Steps of the Demonstration. 1. Prove that the altern. Z BAC = alternate < ACE,

And since AB, CD are not parallels, they will meet. Demon. Again, AB and cd will meet on that side of EF on which are

the Zs which are < 2 right Zs; For suppose them to meet on the other side of EF, as in m,

then KML would be a A ; And since ZS AKL + BKL = 2 right Zs,

13. 1 and Zs CLK + DLK = 2 right Zs, .. the 4 ZS AKL + CLK + BKL + DLK = 4 right Zs, of which ZS BKL + DLK < 2 right Zs, Hypoth. i. ZS AKL + CLK > 2 right Zs, and these,

on the above supposition,
are 2 Zs of the A KML, which is impossible; 17. 1

and .. the supposition is false;
:. AB, CD do not meet towards A and c;
but it has been proved that they will meet,
.. they must meet towards the points B and D;

Wherefore, if a right line, &c. &c. Q. E. D.



2. Prove that exterior ZECD = int. &



ABC, 3. that whole ex. LACD = 2 int. & opp. ZS,

CAB + ABC, 4. that ZS ACD + ACB = Ls CAB + ABC + BCA, 5. that Zs ABC + BCA + CAB = 2 rt. Ls.

Obs. The two corollaries which follow this proposition are of great use, and therefore necessary to be learned; there is one point in the first which may appear strange to a beginner, viz., that the Zs at the

vertex F, are said to be equal

to 4 right Zs; but by inM.

specting the annexed figure he will perceive that the ZS above the straight line mn

are together equal to 2 right Ls (by Prop. XIII. 1.), and also the Zs below mn are together equal to 2 right Zs (by the same Proposition.)

Hence all the ZS EFM + DFE + CFD + NFC + BFN + AFB + MFA = 4 right ZS,

and since | EFM + ZMFA = whole / EFA, and

CFN + 2 BFN = whole _ CFB, .. when the line mn is removed, the <s round the point F will still be together = 4rt. Zs.



Theorem. The straight lines which join the extremities of equal and parallel straight lines towards the same parts are also themselves equal and parallel.

Steps of the Demonstration. 1. Prove that

2 ABC = _ BCD, 2. that (in A S ABC, BCD) base ac = base BD, and

Z ACB = L CBD, that ac || BD.




Theorem. The opposite sides and angles of paral

lelograms are equal to one another, and the diameter bisects them; that is, divides them into two equal parts.

Steps of the Demonstration. 1. Prove that in AS ABC, BCD, ZS ABC, BCA Zs

BCD, CBD, ea. to ea., the adjacent side

BC being common, 2. that :. AB, AC = CD, DB each to each, and _

CAB = BDC, 3. that whole _ ABD = whole / DCA, 4.

that A ABC = A BCD

PROPOSITION XXXV. Theorem. Parallelograms upon the same base, and between the same parallels, are equal to each other.

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