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HERON OF ALEXANDRIA
triangle which is constructed to be constructed on the same
side of the common side as the given triangle is.
A third class (c) is that which avoids reductio ad absurdum,
e.g. a direct proof of I. 19 (for which he requires and gives
a preliminary lemma) and of I. 25.
(4) Heron supplies certain converses of Euclid’s propositions
e.g. of II. 12, 13 and VIII. 27.
(5) A few additions to, and extensions of, Euclid’s propositions
are also found. Some are unimportant, e. g. the construction
of isosceles and scalene triangles in a note on I. 1 and the
construction of hvo tangents in III. 17. The most important
extension is that of III. 20 to the case where the angle at the
circumference is greater than a right angle, which gives an
easy way of proving the theorem of III. 22. Interesting also
are the notes on I. 37 (on I. 24 in Proclus), where Heron
proves that two triangles with two sides of the one equal
to two sides of the other and with the included angles supple
mentary are equal in area, and compares the areas where the
sum of the included angles (one being supposed greater than
the other) is less or greater than two right angles, and on I. 47,
where there is a proof (depending on preliminary lemmas) of
the fact that, in the figure of Euclid’s proposition (see next
page), the straight lines AL, BG, GE meet in a point. This
last proof is worth giving. First come the lemmas.
(1) If in a triangle ABC a straight line DE be drawn
parallel to the base BG cutting the sides AB, AC or those
sides produced in D, E, and if F be the
middle point of BG, then the straight line
AF (produced if necessary) will also bisect
DE. (UK is drawn through A parallel to
DE, and HDL, KEM through D, E parallel
to A F meeting the base in L, M respec
tively. Then the triangles ABF, AFG
between the same parallels are equal. So are the triangles
DBF, EFC. Therefore the differences, the triangles ADF,
AEF, are equal and so therefore are the parallelograms HF,
KF. Therefore LF = FM, or DG = GE.)
(2) is the converse of Eucl. I. 43. If a parallelogram is