3. If the vertex D of the second triangle fall on the line BC, it is evident that BC and BD are unequal. Questions for Examination. 1. What use is made of Prop. vii.? Ans. As a lemma to Prop. viii. 2. In the demonstration of Prop. vii. the contrapositive of Prop. v. occurs; show where. 3. Show that two circles can intersect each other only in one point on the same side of the line joining their centres, and hence that two circles cannot have more than two points of intersection. PROP. VIII.—Theorem. If two triangles (ABC, DEF) have two sides (AB, AC) of one respectively equal to two sides (DE, DF) of the other, and have also the base (BC) of one equal to the base (EF) of the other; then the two triangles shall be equal, and the angles of one shall be respectively equal to the angles of the other—namely, those shall be equal to which the equal sides are opposite. Dem.—Let the triangle ABC be applied to DEF, so that the point B will coincide with E, and the line BC with the line EF; then because BC is equal to EF, the point C shall coincide with F. Then if the vertex A fall on the same side of EF as the vertex D, the point A must coincide with D; for if not, let it take a different position G; then we have EG equal to BA, and BA is equal to ED (hyp.). Hence (Axiom i.) EG is equal to ED: in like manner, FG is equal to FD, and this is impossible [vii.]. Hence the point A must coincide with D, and the triangle ABC agrees in every respect with the triangle DEF; and therefore the three angles of one are respectively equal to the three angles of the other—namely, A to D, B to E, and C to F, and the two triangles are equal. This Proposition is the converse of iv., and is the second case of the congruence of triangles in the Elements. Philo’s Proof.—Let the equal bases be applied as in the foregoing proof, but let the vertices be on the opposite sides; then let BGC be the position which EDF takes. Join AG. Then because BG = BA, the angle BAG = BGA. In like manner the angle CAG = CGA. Hence the whole angle BAC = BGC; but BGC = EDF therefore BAC = EDF. PROP. IX.—Problem. To bisect a given rectilineal angle (BAC). Sol.—In AB take any point D, and cut off [iii.] AE equal to AD. Join DE (Post. i.), and upon it, on the side remote from A, describe the equilateral triangle DEF [i.] Join AF. AF bisects the given angle BAC. Dem.—The triangles DAF, EAF have the side AD equal to AE (const.) and AF common; therefore the two sides DA, AF are respectively equal to EA, AF, and the base DF is equal to the base EF, because they are the sides of an equilateral triangle (Def. xxi.). Therefore [viii.] the angle DAF is equal to the angle EAF; hence the angle BAC is bisected by the line AF. Cor.—The line AF is an axis of symmetry of the figure. Questions for Examination. 1. Why does Euclid describe the equilateral triangle on the side remote from A?