## Homework Assignment 3

### pg.50 prob. 3

Suppose that A,B, and C are noncollinear points in an affine plane. Prove that there is a unique point D such that (A,B,C,D) is a parallelogram.

Construct the line *l* joining A and B, and the line *m* joining B and C. Since A,B and C are noncollinear, the lines *l* and *m* are distinct. Since C is not on *l*, we can construct the unique line *k*, through C and parallel to *l*. Since A is not on *m*, we can construct the unique line *n*, through A and parallel to *m*. Since *n* intersects one of the parallel lines *l* and *k*, it must intersect the other. Let the point of intersection of *n* and *k* be D. (A,B,C,D) is clearly a parallelogram. Now suppose that (A,B,C,D') is a parallelogram. Since the line joining A and D' is parallel to the line joining B and C (i.e. *m*), it must be *n*, since there is a unique line through A parallel to *m*. Thus, D' is on *n*. Similarly, we can show that D' must be on *k*. Hence, D = D' since *n* and *k* can intersect in only one point.

### pg.54 prob.6

Make sketches of the following configurations:
- (A,B,C,D) is a quadrilateral with no pair of its sides parallel.
- In the configuration from (a) insert A' so that (A',B,C,D) is a parallelogram.
- In the configuration from (b) insert B' so that (A,B',C,D) is a parallelogram.
- Similarly, insert C' and D'.

### pg.57 prob.3

Complete the addition table for (OIDF). What is the inverse of D?

+ | O | I |
D | F |
---|

O | O | I | D | F |

I | I | O | F | D |

D | D | F | O | I |

F | F | D | I | O |

The additive inverse of D is D.

### pg.57 prob.6

Suppose that (*P,L*) is a Desarguesian affine plane in which the diagonals of every parallelogram are parallel. Suppose that A is any point in *P*. Prove that relative to any O, A+A = O.

Let A and O be any two points in the Desarguesian affine plane. If A = O, the statement is true, so we may assume that these points are distinct. Let *l* be the line determined by A and O. Choose B, any point not on *l* and draw the line *m* through B parallel to *l*. Let *k* be the line joining O and B and *n* the line through A parallel to *k*. The lines *l* and *n* meet at a point D. Notice that (BOAD) is a parallelogram. To determine A+A, we need to construct the line parallel to the line determined by A and B which passes through D. Since the line determined by A and B is a diagonal of the parallelogram, the unique line through D which is parallel to it is the other diagonal (by hypothesis), i.e. the line joining O and D. Since this line meets *l* at O, we have A+A = O.

### pg.61 prob.3

Let (A,B,C,D) be a quadrilateral in a Desarguesian affine plane, none of whose sides are parallel. Define A' to be the unique point such that (A',B,C,D) is a parallelogram, and B' the unique point such that (A,B',C,D) is a parallelogram. Prove that the line joining A and A' is parallel to the line joining B and B'.

By construction, lines *l*(C,D) and *l*(A,B') are parallel, as are the lines *l*(C,D) and *l*(A',B). Consider the triangles DAA' and CB'B. The lines joining corresponding vertices of these triangles are mutually parallel. Also by construction, lines *l*(A,D) and *l*(C,B') are parallel, as are *l*(D,A') and *l*(C,B). By Desargues theorem, the third sides of these triangles, namely *l*(A,A') and *l*(B',B) are parallel.

### pg.65 prob.4

Let O, I and A be distinct collinear points in a Desarguesian affine plane. Define the point 2I to be I + I and 2A to be the point A + A, the addition done relative to O. Prove geometrically that (2I)A = A + A assuming that the points O, I, 2I, A and 2A are all distinct.

Choose point B and construct 2I.

Using B, construct (2I)A.

Again using B, construct 2A.

To prove that (2I)A = A + A = 2A, we must show that E = E'. Consider triangles DD'E' and IBA. Since, by construction, *l*(B,A) || *l*(D',E'), *l*(I,B) || *l*(D,D') and *l*(D,E') || *l*(I,A) (i.e., the corresponding sides of the triangles are parallel in pairs) and *l*(D',B) || *l*(I,D) we have by the converse of Desargues theorem (affine version) that *l*(A,E') || *l*(I,D || *l*(O,B). Thus, *l*(A,E') = *l*(A,E) and the conclusion follows.