VI. 19, /SABG : IS ABD = AG' : AD\

Also the triangles ABG, ABD are of equal altitude; hence, by Euc. VI. 1,
l\ABG'. AABD = BG:BD,
.-. AG':AD'=BG : BD.
. AG^ _AD' •'• BG ~ BD •
* See a note by M, Ooccoz in U Illustration, Paris, Jan. 12, 1895.
CH. Ill] GEOMETRICAL RECREATIONS 47
Hence, by Enc. ii. 13,
AB' + BG^ -2BG.be ^ AB'^BD'-2BD . BE BG ~ BD '
.\^+BG-2BE=^~ + BD- 2BE, . ^^' BD-^^' EC
• ^^^ - B^ '^^ _ ^B^ - BG . BD BG BD •
.\ BG = BD,
a result which is impossible.
Third Fallacy*. To prove that the sum of the lengths of two sides of any triangle is equal to the length of the third side.
A^ , , , ,D
Let ABG be a triangle. Complete the parallelogram of which AB and BG are sides. Divide AB into n + 1 equal parts, and through the points so determined draw n lines parallel to BG. Similarly, divide BG into n + 1 equal parts, and through the points so determined draw n lines parallel to AB. The parallelogram ABGD is thus divided into (n + 1)^ equal and similar parallelograms.
I draw the figure for the case in which n is equal to 3, then, taking the parallelograms of which ^C is a diagonal, as indicated in the diagram, we have
AB + BG^AG + HJ + KL + MN
+ GH-^JK + LM + NG.
A similar relation is true however large n may be. Now let n increase indefinitely. Then the lines AG, GH, &c. will
* The Canterbury Puzzles, by H. E. Dudeney, London, 1907, pp. 26—28.
48 GEOMETRICAL RECREATIONS [CH. Ill
get smaller and smaller. Finally the points (r, J, i, ... will approach indefinitely near the line AG, and ultimately will lie on it; when this is the case the sum of AG and GH will be equal to AH, and similarly for the other similar pairs of lines. Thus, ultimately,
AB + BC^AH + HK + KM + MG
= AG,
a result which is impossible.
Fourth Fallacy. To prove that every triangle is isosceles. Let ABG be any triangle. Bisect BG in D, and through D draw DO perpendicular to BG. Bisect the angle BAG by AO.
First. If DO and ^0 do not meet, then they are parallel. Therefore ^10 is at ri-ht angles to BG. Therefore AB = AG.
Second. If DO and AO meet, let them meet in 0. Draw OE perpendicular to AG. Draw OF perpendicular to AB. Join OB, OG.
Let us begin by taking the case where 0 is inside the triangle, in which case E falls on ^C and F on BG.
The triangles AOF and AOE are equal, since the side ^0 is common,
angle OAF= angle OAE, and angle OF A = angle OEA. Hence AF=AE. Also, the triangles BOF axid GOE are equal. For since OD bisects BG at right angles, we have OB = 00] also, since the triangles AOF and AOE are equal, we have 0F= OE; lastly, the angles at F and E are right angles. Therefoie, by Euc. I. 47 and I. 8, the triangles BOF and GOE are equal. Hence FB = EG.
Therefore AF + FB ^ AE ■{■ EG, that is, AB = AG.
The same demonstration will cover the case where DO and AO meet at D, as also the case where they meet outside BG but so near it that E and F fall on AG and AB and not on AG and AB produced.
Next take the case where DO and AO meet outside the triangle, and E and F fall on AG and AB produced. Draw
CH. Ill]
GEOMETRICAL RECREATIONS
49
OE perpenrlionlar to AC produced. Draw OF perpendicular to ^^ produced. Join OB, 00.
Following the same argument as before, from the equality of the triangles AOF and AOE, we obtain AF=AE\ and, from the equality of the triangles BOF and GOE, we obtain FB = EG. Therefore AF-FB = AE-EG, that is, AB = A G.
Thus in all cases, w^hether or not DO and AO meet, and whether they meet inside or outside the triangle, we have AB=AG: and therefore every triangle is isosceles, a result which is impossible.
Fifth Fallacy'*. To prove that 'ttJ^ is equal to ir/S. On the hypothenuse, BO, of an isosceles right-angled triangle, DBG, describe an equilateral triangle ABO, the vertex A being on the same side of the base as D is. On QA take a point H so that OH = GD. Bisect BD in K. Join HK and let it cut OB (produced) in L. Join DL. Bisect DL at 31, and through M draw MO perpendicular to DL. Bisect HL at N, and through N draw NO perpendicular to HL. Since DL and HL intersect, therefore MO and NO will also intersect ; moreover, since BDG is a right angle, MO and NO both slope away from DG and therefore they will meet on the side of DL remote from A. Join 00, OD, OH, OL.
The triangles OMD and OML are equal, hence OD = OL. Similarly the triangles ONL and ONH are equal, hence OL = OH. Therefore OD = OH. Now in the triangles OCD and OGH, we have OD = OH, GD = GH (by construction), and
* This ingenious fallacy is due to Captain Turton : it appeared for the first time in the third edition of this work.
B. R.
4
50
GEOMETRICAL RECREATIONS
[CH. HI
OG common, hence (by Euc. I. 8) the angle OCD is equal to the angle OGH. Hence the angle BGD is equal to the angle BGH, that is, 7r/4 is equal to 7r/3, which is absurd.
Sixth Fallacy*. To prove that, if two opposite sides of a quadrilateral are equal, the other two sides must he parallel. Let A BGD be a quadrilateral such that AB is equal to DG. Bisect AD in M, and through M draw MO at right angles to AD. Bisect BG in N, and draw NO at right angles to BG.
If MO and NO are parallel, then AD and BG (which are at right angles to them) are also parallel.
If MO and NO are not parallel, let them meet in 0 ; then 0 must be either inside the quadrilateral as in the left-hand
diagram or outside the quadrilateral as in the right-hand diagram. Join OA, OB, OG, OD.
Since OM bisects AD and is perpendicular to it, we have OA = OD, and the angle 0AM equal to the angle ODM. Similarly OB = 00, and the angle OBN is equal to the angle OGN. Also by hypothesis AB = DG, hence, by Euc. i. 8, the triangles OAB and ODG are equal in all respects, and therefore the angle AOB is equal to the angle DOG,
Hence in the left-hand diagram the sum of the angles AOM, AOB is equal to the sum of the angles DOM, DOG; and in the right-hand diagram the difference of the angles AOM,AOB is equal to the difference of the angles DOM, DOG; and therefore in both cases the angle MOB is equal to the angle MOG, i.e. OM (or OM produced) bisects the angle BOG, But the angle NOB is equal to the angle NOG, i.e. ON bisects the angle BOG] hence OM and ON coincide in directioa
* i^lathesis, October, 1893, series 2, vol. in, p. 224.
CH. Ill] GEOMETRICAL KECREATIONS 51
Therefore AD and DC, which are perpendicular to this direc- tion, must be parallel. This result is not uuiversally true, and the above demonstration contains a flaw.
Seventh Fallacy. The following argument is taken from a text-book on electricity, published in 1889 by two distinguished mathematicians, in which it was presented as valid. A given vector OP of length I can be resolved in an infinite number of ways into two vectors OM, MP, of lengths l\ I", and we can make Vjl" have any value we please from nothing to infinity. Suppose that the system is referred to rectangular axes Ox, Oy\ and that OP, OM, MP make respectively angles 6, 6'y 6" with Ox. Hence, by projection on Oy and on Ox,
we have
Z sin (9 = V sin 6' + I" sin 6\
Zcos(9 = rcos^'-}-rcos6'".
n sin 9' -I- sin 6"
\ tan^ =
• •
ncos6' + cos 0
/' >
where n^Vjl". This result is true whatever be the value of n. But n may have any value {ex. gr. ?i=x, or n = 0), hence tan 6 — tan 6' = tan 9"^ which obviously is impossible.
Eighth Fallacy*. Here is a fallacious investigation of the value of tt: it is founded on well-known quadratures. The area of the semi-ellipse bounded by the minor axis is (in the usual notation) equal to ^irab. If the centre is moved off to an indefinitely great distance along the major axis, the ellipse degenerates into a parabola, and therefore in this particular limiting position the area is equal to two-thirds of the circum- scribing rectangle. But the first result is true whatever be the dimensions of the curve.
.'. ^7rah= |a X 26,
.-. 7r=8/3,
a result which obviously is untrue.
Ninth Fallacy. Ever-y ellipse is a circle. The focal distance of a point on an ellipse is given (in the usual notation) in terms
* Tliis was communicated to me hy Mr 11, Chai ties.
1 o
52 GEOMETRICAL RECREATIONS [CH. Ill
of the abscissa by the formula r=a+ex. Hence dr/dx=e. From this it follows that r cannot have a maximum or minimum value. But the only closed curve in which the radius vector has not a maximum or minimum value is a circle. Hence, every ellipse is a circle, a result which obviously is untrue.
Geometrical Paradoxes. To the above examples I may add the following questions, which, though not exactly falla- cious, lead to results which at a hasty glance appear impossible.
First Paradox. The first is a problem, sent to me by Mr W. Ronton, to rotate a plane lamina (say, for instance, a sheet of paper) through four right angles so that the effect is equivalent to turning it through only one right angle.
Second Paradox. As in arithmetic, so in geometry, the theory of probability lends itself to numerous paradoxes. Here is a very simple illustration. A. stick is broken at random into three pieces. It is possible to put them together into the shape of a triangle provided the length of the longest piece is less than the sum of the other two pieces {cf. Euc. L 20), that is, provided the length of the longest piece is less than half the length of the stick. But the probability that a fragment of a stick shall be half^ the original length of the stick is 1/2. Hence the probability that a triangle can be constructed out of the three pieces into which the stick is broken would appear to be 1/2. This is not true, for actually the probability is 1/4.
Third Paradox. The following example illustrates how easily the eye may be deceived in demonstrations obtained by actually dissecting the figures and re-arranging the parts. In fact proofs by superposition should be regarded with consider- able distrust unless they are supplemented by mathematical reasoning. The well-known proofs of the propositions Euclid