The First Six Books of the Elements of Euclid — John Shaqi
The First Six Books of the Elements of EuclidEuclid
Science
The First Six Books of the Elements of Euclid
Euclid
Euclid's Elements; Mathematics, Greek
If two triangles (ABC, DEF) have two sides (AB, AC) of one respectively equal
to two sides (DE, DF) of the other, but the base (BC) of one greater than the base
(EF) of the other, the angle (A) contained by the sides of that which has
the greater base is greater them the angle (D) contained by the sides of the
other.
Dem.—If the angle A be not greater than D, it must be either equal to it or less
than it. We shall examine each case:—
1. If A were equal to D, the triangles ABC, DEF would have the two sides AB,
AC of one respectively equal to the two sides DE, DF of the other, and the angle A
contained by the two sides of one equal to the angle D contained by the two
sides of the other. Hence [iv.] BC would be equal to EF; but BC is, by
hypothesis, greater than EF; hence the angle A is not equal to the angle
D.
2. If A were less than D, then D would be greater than A, and the triangles
DEF, ABC would have the two sides DE, DF of one respectively equal to the two
sides AB, AC of the other, and the angle D contained by the two sides of one greater
than the angle A contained by the two sides of the other. Hence [xxiv.] EF would be
greater than BC; but EF (hyp.) is not greater than BC. Therefore A is not less
than D, and we have proved that it is not equal to it; therefore it must be
greater.
Or thus, directly: Construct the triangle ACG, whose three sides AG, GC, CA
shall be respectively equal to the three sides DE, EF, FD of the triangle DEF
[xxii.]. Join BG. Then because BC is greater than EF, BC is greater than CG.
Hence [xviii.] the angle BGC is greater than GBC; and make (xxiii.) the angle
BGH equal to GBH, and join AH. Then [vi.] BH is equal to GH. Therefore the
triangles ABH, AGH have the sides AB, AH of one equal to the sides AG, AH of
the other, and the base BH equal to GH. Therefore [viii.] the angle BAH is equal to
GAH. Hence the angle BAC is greater than CAG, and therefore greater than
EDF.
Exercise.
Demonstrate this Proposition directly by cutting off from BC a part equal to EF.
PROP. XXVI.—Theorem.
If two triangles (ABC, DEF) have two angles (B, C) of one equal respectively to
two angles (E, F) of the other, and a side of one equal to a side similarly placed
with respect to the equal angles of the other, the triangles are equal in every
respect.
Dem.—This Proposition breaks up into two according as the sides given to be
equal are the sides adjacent to the equal angles, namely BC and EF, or those
opposite equal angles.
Public-domain text, read in full here on John Shaqi.
Reviews
Reviews
No reviews yet
Be the first to share your thoughts on this work.
Join the Discussion
Join the discussion
Sign in to leave a comment or review.
Sign InorCreate an account