QUESTIONS AND EXERCISES

1. Prove that if one angle of a parallelogram be a right angle, all its angles are right angles.

2. Prove that if the opposite angles of a quadrilateral figure be equal to one another it is a parallelogram.

3. Prove that the diagonals of a parallelogram bisect each other.

4. Make a rectangle which shall be equal in area to a given parallelogram.

5. Make a rhombus which shall be equal in area to a given parallelogram.

6. If a square and any other parallelogram of equal area stand on the same base, the perimeter of the square shall be less than that of the parallelogram.

Lesson No. 17

PROPOSITION 38. THEOREM

Triangles on equal bases, and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD: then the triangle ABC shall be equal to the triangle DEF.

Construction: Produce AD both ways to the points G, H; through B draw BG parallel to CA, and through F draw FH parallel to ED (Proposition 31).

Proof: Each of the figures GBCA and DEFH is a paral lelogram (definition). And they are equal to one another,

because they are on equal bases BC, EF, and between the same parallels, BF and GH (Proposition 36). And the triangle ABC is half of the parallelogram GBCA, because the diagonal AB bisects it (Proposition 34); and the triangle DEF is half of the parallelogram DEFH, because the diagonal DF bisects it. But the halves of equal things are equal. Therefore the triangle ABC is equal to the triangle DEF.

PROPOSITION 39. THEOREM

Equal triangles on the same base, and on the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC be on the same base BC, and on the same side of it: then they shall be between the same parallels.

B

Construction: Join AD.

Proof: AD shall be parallel to BC. For if it is not, through A draw AE parallel to BC (Proposition 31) and join EC.

The triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE (Proposition 37). But the triangle ABC is given equal to the triangle DBC. Therefore also the triangle DBC is equal to the triangle EBC, the greater to the less, which is impossible. Therefore AE is not parallel to BC.

In the same manner it can be demonstrated, that no other line through A but AD is parallel to BC. Therefore AD is parallel to BC.

QUESTIONS AND EXERCISES

1. Prove Proposition 38 when the vertices of the triangles fall upon the same point.

2. Prove that the four triangles into which a parallelogram is divided by its diagonals are equal to one another.

3. Prove that the straight line which joins the middle points of two sides of a triangle is parallel to the base.

Equal triangles, on equal bases in the same straight line, and toward the same parts, are between the same parallels.

Let the equal triangles ABC and DEF be on equal bases BC, EF, in the same straight line BF, and toward the same parts: then they shall be between the same parallels.

Proof: AD shall be parallel to BF. For if it is not, through A draw AG parallel to BF (Proposition 31) and join GF.

The triangle ABC is equal to the triangle GEF, because they are on equal bases BC, EF, and between the same parallels (Proposition 38). But the triangle ABC is given equal to the triangle DEF. Therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible. Therefore AG is not parallel to BF.

In the same manner it can be demonstrated that no other line through A but AD is parallel to BF. Therefore AD is parallel to BF.

PROPOSITION 41. THEOREM

If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle.

Let the parallelogram ABCD and the triangle EBC be on the same base BC, and between the same parallels BC, AE: then the parallelogram ABCD shall be double of the triangle EBC.

Proof: The triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE (Proposition 37). But the parallelogram ABCD is double of the triangle ABC, because the diagonal AC bisects the parallelogram (Proposition 34). Therefore the parallelogram ABCD is also double of the triangle EBC.

QUESTIONS AND EXERCISES

1. Prove that if two triangles or two parallelograms be upon equal bases and have the same altitude, they shall be equal to one another.

2. Prove that the straight line joining the middle points of two sides of a triangle is parallel to the third side..

3. Prove that the line drawn parallel to any side of a triangle through the middle point of another side bisects: the third side.

Lesson No. 19

PROPOSITION 42. PROBLEM

To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle: it is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

Construction: Bisect BC in E (Proposition 10). Join AE, and at the point E in the straight line EC make the angle CEF equal to D (Proposition 23). Through A draw AFG parallel to EC, and through C draw CG parallel to EF (Proposition 31). Then FECG is a parallelogram (Definition 11).

Proof: Because BE is equal to EC the triangle ABE is equal to the triangle AEC, for they are on equal bases BE, EC, and between the same parallels BC, AG (Proposition 38). Therefore the triangle ABC is double the triangle AEC. But the parallelogram FECG is also double the triangle AEC, because they are on the same base EC, and between the same parallels EC, AG (Proposition 41). Therefore the parallelogram FECG is equal to the triangle ABC; and it has one of its angles CEF equal to the given angle D.

Therefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D.