2. Prove that the angle BCA is greater than EFD.
PROP. XXV.—Theorem.
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.