Commentary on Proposition VI.16¶
Back to Proposition VI.16.
This proposition is a particular case of VI. 14, but one which is on all accounts worth separate statement. It may also be enunciated in the following form:
Rectangles which have their bases reciprocally proportional to their heights are equal in area; and equal rectangles have their bases reciprocally proportional to their heights.
Since any parallelogram is equal to a rectangle of the same height and on the same base, and any triangle with the same height and on the same base is equal to half the parallelogram or rectangle, it follows that Equal parallelograms or triangles have their bases reciprocally proportional to their heights and vice versa.
The present place is suitable for giving certain important propositions, including those which Simson adds to Book VI. as Props. B, C and D, which are proved directly by means of VI. 16.
Proposition B is a particular case of the following theorem.
If a circle be circumscribed about a triangle :g:`ABC <0229.g1:< A B C >>` and there be drawn through :g:`A <0229.g1:A>` any two straight lines either both within or both without the angle :g:`BAC <0229.g1:< B A C )>`, viz.
Join CE.
The angles BAD, EAC are equal, by hypothesis; and the angles ABD, AEC are equal. [III. 21, 22]
Hence
There are now two particular cases to be considered.
(a) Suppose that AD, AE coincide; ADE will then bisect the angle BAC.
(b) Suppose that AD, AE are in one straight line but that D, E are on opposite sides of A; AD will then bisect the external angle at A.
In the first case (a) we have
The following converse of Simson’s Prop. B may be given: If a straight line :g:`AD <0230.g1:/ A D />` be drawn from the vertex :g:`A <0230.g1:A>` of a triangle to meet the base, so that the square on :g:`AD <0230.g1:/ A D />` together with the rectangle :g:`BD <0230.g1:/ B D />`, :g:`DC <0230.g1:/ D C />` is equal to the rectangle :g:`BA <0230.g1:/ B A />`, :g:`AC <0230.g1:/ A C />`, the line :g:`AD <0230.g1:/ A D />` will bisect the angle :g:`BAC <0230.g1:< B A C )>` except when the sides :g:`AB <0230.g1:/ A B />`, :g:`AC <0230.g1:/ A C />` are :eq:`equal <0230.g1:/ A B / = 0230.g1:/ A C />`, in which case every line drawn to the base will have the property mentioned.
Let the circumscribed circle be drawn, and let AD produced meet it in E; join CE.
Add to each the square on AD;
Hence
(a) If they are equal, the angles BAD, EAC are also equal, and AD bisects the angle BAC.
(b) If they are supplementary, the angle ADC must be equal to the angle ACE.
Therefore the angles BAD, ABD are together equal to the angles ACB, BCE, i.e. to the angles ACD, BAD.
Take away the common angle BAD, and
Euclid himself assumes, in Prop. 67 of the $Data$, the result of so much of this proposition as relates to the case where BA = AC. He assumes namely, without proof, that, if BA = AC, and if D be any point on BC, the rectangle BD, DC together with the square on AD is equal to the square on AB.
PROPOSITION C.¶
Let ABC be a triangle and AD the perpendicular on BC. Draw the diameter AE of the circle circumscribed about the triangle ABC.
Join EC.
Since the right angle BDA is equal to the right angle ECA in a semicircle, [III. 31]
and the angles ABD, AEC in the same segment are equal, [III. 21]
Therefore,
This result corresponds to the trigonometrical formula for $R$, the radius of the circumscribed circle,
$$R = \frac{abc}{4\Delta}.$$
PROPOSITION D.¶
This is the highly important lemma given by Ptolemy (ed. Heiberg, Vol. I, pp. 36—7) which is the basis of his calculation of the table of chords in the section of Book I. of the μεγάλη σύνταξις entitled “concerning the size of the straight lines [i.e. chords] in the circle” (περὶ τῆς πηλικότητος τῶν ἐν τῷ κύκλῳ εὐθειῶν).
The theorem may be enunciated thus.
I shall give the proof in Ptolemy’s words, with the addition only, in brackets, of two words applying to a second figure not given by Ptolemy.
“Let there be a circle with any quadrilateral ABCD inscribed in it, and let AC, BD be joined.
It is to be proved that the rectangle contained by AC and BD is equal to the sum of the rectangles AB, DC and AD, BC.
For let the angle ABE be made equal to the angle contained by DB, BC.
If then we add [or subtract] the angle EBD,
Hence, proportionally,
Again, since the angle ABE is equal to the angle DBC, and the angle BAE is also equal to the angle BDC, [III. 21]
Therefore, proportionally,
But it was also proved that
Another proof of this proposition, and of its converse, is indicated by Dr Lachlan (Elements of Euclid, pp. 273—4). It depends on two preliminary propositions.
If two circles be divided, by a chord in each, into segments which are similar respectively, the chords are proportional to the corresponding diameters.
The proof is instantaneous if we join the ends of each chord to the centre of the circle which it divides, when we obtain two similar triangles.
If :g:`D <0232.g2:D>` be any point on the circle circumscribed about a triangle :g:`ABC <0232.g2:< A B C >>`, and :g:`DX <0232.g2:/ D X />`, :g:`DY <0232.g2:/ D Y />`, :g:`DZ <0232.g2:/ D Z />` be perpendicular to the sides :g:`BC <0232.g2:/ B C />`, :g:`CA <0232.g2:/ C A />`, :g:`AB <0232.g2:/ A B />` of the triangle respectively, then :g:`X <0232.g2:X>`, :g:`Y <0232.g2:Y>`, :g:`Z <0232.g2:Z>` lie in one straight line; and, conversely, if the feet of the perpendiculars from any point :g:`D <0232.g2:D>` on the sides of a triangle lie in one straight line, :g:`D <0232.g2:D>` lies on the circle circumscribed about the triangle.
The proof depending on III. 21, 22 is well known.
Now suppose that D is any point in the plane of a triangle ABC, and that DX, DY, DZ are perpendicular to the sides BC, CA, AB respectively.
Join YZ, DA.
Then, since the angles at Y, Z are right, A, Y, D, Z lie on a circle of which DA is the diameter.
And YZ divides this circle into segments which are similar respectively to the segments into which BC divides the circle circumscribing ABC, since the angles ZAY, BAC coincide, and their supplements are equal.
Therefore, if d be the diameter of the circle circumscribing ABC,
Hence, in a quadrilateral in general, the rectangle contained by the diagonals is less than the sum of the rectangles contained by the pairs of opposite sides.
Next, suppose that D lies on the circle circumscribed about ABC, but so that A, B, C, D follow each other on the circle in this order, as in the figure annexed.
Let DX, DY, DZ be perpendicular to BC, CA, AB respectively, so that X, Y, Z are in a straight line.
Then, since the rectangles AD, BC; BD, CA; CD, AB are equal to the rectangles d, YZ; d, ZX; d, XY respectively, and XZ is equal to the sum of XY, YZ, so that the rectangle d, XZ is equal to the sum of the rectangles d, XY and d, YZ, it follows that
$Conversely$, if the latter statement is true, while we are supposed to know nothing about the position of D, it follows that
so that X, Y, Z must be in a straight line.
Hence, from the theorem (2) above, it follows that D must lie on the circle circumscribed about ABC, i.e. that ABCD is a quadrilateral about which a circle can be described.
All the above propositions can be proved on the basis of Book III. and without using Book VI., since it is possible by the aid of III. 21 and 35 alone to prove that in equiangular triangles the rectangles contained by the non-corresponding sides about equal angles are equal to one another (a result arrived at by combining VI. 4 and VI. 16). This is the method adopted by Casey, H. M. Taylor, and Lachlan; but I fail to see any particular advantage in it.
PROPOSITION E.¶
Lastly, the following proposition may be given which Playfair added as VI. E. It appears in the $Data$ of Euclid, Prop. 93, and may be thus enunciated.
Join CD. Then, since AD bisects the angle BAC, the subtended arcs BD, DC, and therefore the chords BD, DC, are equal.
The result can now be easily deduced from Ptolemy’s theorem.
For the rectangle AD, BC is equal to the sum of the rectangles AB, DC and AC, BD, i.e. (since BD, CD are equal) to the rectangle contained by BA + AC and BD.
Euclid proves it differently in $Data$, Prop. 93.
Let AD meet BC in E.
Then, since AE bisects the angle BAC,
or, alternately,
Therefore also
Again, since the angles BAD, EAC are equal, and the angles ADB, ACE are also equal, [III. 21]
Therefore
Hence
and, alternately,
Euclid concludes that, if the circle ABC is given in magnitude, and the chord BC cuts off a segment of it containing a given angle (so that, by $Data$ Prop. 87, BC and also BD are given in magnitude),
and further that (since, by similar triangles, BD is to DE as AC is to CE, while BA + AC is to BC as AC is to CE),