2. Prove that the line joining the point A to the intersection of the lines CF and BG is an axis of symmetry of the figure. 3. If two isosceles triangles be on the same base, and be either at the same or at opposite sides of it, the line joining their vertices is an axis of symmetry of the figure formed by them. 4. Show how to prove this Proposition by assuming as an axiom that every angle has a bisector. 5. Each diagonal of a lozenge is an axis of symmetry of the lozenge. 6. If three points be taken on the sides of an equilateral triangle, namely, one on each side, at equal distances from the angles, the lines joining them form a new equilateral triangle. PROP. VI.—Theorem. If two angles (B, C) of a triangle be equal, the sides (AC, AB) opposite to them are also equal. Dem.—If AB, AC are not equal, one must be greater than the other. Suppose AB is the greater, and that the part BD is equal to AC. Join CD (Post. i.). Then the two triangles DBC, ACB have BD equal to AC, and BC common to both. Therefore the two sides DB, BC in one are equal to the two sides AC, CB in the other; and the angle DBC in one is equal to the angle ACB in the other (hyp). Therefore [iv.] the triangle DBC is equal to the triangle ACB—the less to the greater, which is absurd; hence AC, AB are not unequal, that is, they are equal. Questions for Examination. 1. What is the hypothesis in this Proposition? 2. What Proposition is this the converse of? 3. What is the obverse of this Proposition? 4. What is the obverse of Prop. v.? 5. What is meant by an indirect proof? 6. How does Euclid generally prove converse Propositions? 7. What false assumption is made in the demonstration? 8. What does this assumption lead to? PROP. VII—Theorem. If two triangles (ACB, ADB) on the same base (AB) and on the same side of it have one pair of conterminous sides (AC, AD) equal to one another, the other pair of conterminous sides (BC, BD) must be unequal. Dem.—1. Let the vertex of each triangle be without the other. Join CD. Then because AD is equal to AC (hyp.), the triangle ACD is isosceles; therefore [v.] the angle ACD is equal to the angle ADC; but ADC is greater than BDC (Axiom ix.); therefore ACD is greater than BDC: much, more is BCD greater than BDC. Now if the side BD were equal to BC, the angle BCD would be equal to BDC [v.]; but it has been proved to be greater. Hence BD is not equal to BC. 2. Let the vertex of one triangle ADB fall within the other triangle ACB. Produce the sides AC, AD to E and F. Then because AC is equal to AD (hyp.), the triangle ACD is isosceles, and [v.] the external angles ECD, FDC at the other side of the base CD are equal; but ECD is greater than BCD (Axiom ix.). Therefore FDC is greater than BCD: much more is BDC greater than BCD; but if BC were equal to BD, the angle BDC would be equal to BCD [v.]; therefore BC cannot be equal to BD.