EXERCISES ON PROPOSITIONS 37-40.

DEFINITION. Each of the three straight lines which join the angular points of a triangle to the middle points of the opposite sides is called a Median of the triangle.

ON PROP. 37.

1. If, in the figure of Prop. 37, AC and BD intersect in K, shew that (i) the triangles AKB, DKC are equal in area.

(ii) the quadrilaterals EBKA, FCKD are equal.

2. In the figure of 1. 16, shew that the triangles ABC, FBC are equal in area.

3. On the base of a given triangle construct a second triangle, equal in area to the first, and having its vertex in a given straight line.

4.

Describe an isosceles triangle equal in area to a given triangle and standing on the same base.

ON PROP. 38.

5. A triangle is divided by each of its medians into two parts of equal area.

6. A parallelogram is divided by its diagonals into four triangles of equal area.

7. ABC is a triangle, and its base BC is bisected at X; if Y be any point in the median AX, shew that the triangles ABY, ACY are equal in area.

8. In AC, a diagonal of the parallelogram ABCD, any point X is ́taken, and XB, XD are drawn: shew that the triangle BAX is equal to the triangle DAX.

9. If two triangles have two sides of one respectively equal to two sides of the other, and the angles contained by those sides supplementary, the triangles are equal in area.

10.

ON PROP. 39.

The straight line which joins the middle points of two sides of a triangle is parallel to the third side.

11. If two straight lines AB, CD intersect in O, so that the triangle AOC is equal to the triangle DOB, shew that AD and CB are parallel.

ON PROP. 40.

12. Deduce Prop. 40 from Prop. 39 by joining AE, AF in the figure of page 72.

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 upon the same base BC, and between the same parallels BC, AE:

then shall the parallelogram ABCD be double of the triangle

EBC.

Proof. Then the triangle ABC is equal to the triangle EBC, for they are on the same base BC, and between the same parallels BC, AE.

I. 37.

But the parallelogram ABCD is double of the triangle ABC, for the diagonal AC bisects the parallelogram. I. 34. Therefore the parallelogram ABCD is also double of the triangle EBC.

Q.E.D.

1.

EXERCISES.

ABCD is a parallelogram, and X, Y are the middle points of the sides AD, BC; if Z is any point in XY, or XY produced, shew that the triangle AZB is one quarter of the parallelogram ABCD.

2. Describe a right-angled isosceles triangle equal to a given square.

3. If ABCD is a parallelogram, and XY any points in DC and AD respectively: shew that the triangles AXB, BYC are equal in area.

4. ABCD is a parallelogram, and P is any point within it; shew that the sum of the triangles PAB, PCD is equal to half the parallelogram.

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 angle.

Let ABC be the given triangle, and D the given angle. It is required to describe a parallelogram equal to ABC, and having one of its angles equal to D.

At E in CE, make the angle CEF equal to D;
through A draw AFG parallel to EC;
and through C draw CG parallel to EF.
Then FECG shall be the parallelogram required.
Join AE.

I. 10.

I. 23.

I. 31.

Proof. Now the triangles ABE, AEC are equal, for they are on equal bases BE, EC, and between the same parallels; I. 38. therefore the triangle ABC is double of the triangle AEC. But FECG is a parallelogram by construction; Def. 26. and it is double of the triangle AEC,

for they are on the same base EC, and between the same parallels EC and AG.

I. 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.

Q. E. F.

EXERCISES.

1. Describe a parallelogram equal to a given square standing on the same base, and having an angle equal to half a right angle.

2. Describe a rhombus equal to a given parallelogram and standing on the same base. When does the construction fail?

DEFINITION.

If in the diagonal of a parallelogram any point is taken, and straight lines are drawn through it parallel to the sides of the parallelogram; then of the four parallelograms into which the whole figure is divided, the two through which the diagonal passes are called Parallelograms about that diagonal, and the other two, which with these make up the whole figure, are called the complements of the parallelograms about the diagonal.

Thus in the figure given below, AEKH, KGCF are parallelograms about the diagonal AC; and HKFD, EBGK are the complements of those parallelograms.

NOTE. A parallelogram is often named by two letters only, these being placed at opposite angular points.

PROPOSITION 43. THEOREM.

The complements of the parallelograms about the diagonal of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, and KD, KB the complements of the parallelograms EH, GF about the diagonal AC: then shall the complement BK be equal to the comple

ment KD.

Proof. Because EH is a parallelogram, and AK its diagonal, therefore the triangle AEK is equal to the triangle AHK. 1. 34. For a similar reason the triangle KGC is equal to the triangle KFC.

Hence the triangles AEK, KGC are together equal to the triangles AHK, KFC.

1. 34.

But the whole triangle ABC is equal to the whole triangle ADC, for AC bisects the parallelogram ABCD ; therefore the remainder, the complement BK, is equal to the remainder, the complement KD.

Q.E.D.

EXERCISES.

In the figure of Prop. 43, prove that

(i) The parallelogram ED is equal to the parallelogram BH.
(ii) If KB, KD are joined, the triangle AKB is equal to the
triangle AKD.

To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle.

Let AB be the given straight line, C the given triangle, and D the given angle.

It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to the angle D.

Construction. On AB produced describe a parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D; I. 22 and I. 42*. through A draw AH parallel to BG or EF, to meet FG produced in H.

Join HB.

I. 31.

This step of the construction is effected by first describing on AB produced a triangle whose sides are respectively equal to those of the triangle C (1. 22); and by then making a parallelogram equal to the triangle so drawn, and having an angle equal to D (1. 42).