EXERCISES.

I. Bisect a triangle by a line passing through one of its angular points.

2. Any line drawn through the intersection of the diagonals of a parallelogram to meet the sides bisects the figure.

3.

Find the locus of the vertices of triangles of equal area upon the same base.

4. If the sides of a triangle are 3, 4, 5 inches respectively, the triangle is right-angled.

5. Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in that point.

6. The diagonals of a parallelogram divide it into four equivalent triangles.

7. If from any point in the diagonal of a parallelogram. straight lines be drawn to the angles, then the parallelogram will be divided into two pairs of equivalent triangles.

8. ABCD is a parallelogram, and E any point in the diagonal AC produced. Shew that the triangles EBC, EDC will be equivalent.

9. ABCD is a parallelogram, and O any point within it, shew that the triangles OAB, OCD are together equivalent to half the parallelogram.

IO. On the same supposition if lines are drawn through O parallel to the sides of the parallelogram, then the difference of the parallelograms DO, BO is double of the triangle OAC.

II. The diagonals of a parallelogram ABCD intersect in O, and P is a point within the triangle OAB. Prove that the difference of the triangles APB, CPD, is equivalent to the sum of the triangles APC, BPD.

12. If the points of bisection of the sides of a triangle be joined, the triangle so formed shall be one-fourth of the given triangle.

13. Shew that the sum of the squares on the lines joining the angular points of a square to any point within it is double of the sum of the squares on the perpendiculars from that point on the sides.

14. If the sides of a quadrilateral figure be bisected, and the points of bisection joined, prove that the figure so formed will be a parallelogram equal in area to half the given quadrilateral.

15. Bisect a parallelogram by a line passing through any given point.

SECTION II.

PROBLEMS.

On the Quadrature of a Rectilineal Area.

There is one problem which from its historical interest, and from the valuable illustrations it affords of the methods and limitations of Geometry, should find a place there. This problem is called the quadrature of a rectilineal area, which means the finding a square whose area is equivalent to that of any given figure which is bounded by straight lines. It gave a means of comparing any two dissimilarly

shaped rectilineal figures, such as irregularly shaped fields whose boundaries were straight. In the present condition of mathematics it is not necessary, as the student will hereafter learn, but it will always be instructive.

The problem is approached by the following stages:

(1) To construct a parallelogram, with sides inclined at a given angle, equal to a given triangle.

(2) To construct on a given straight line a parallelogram, with sides inclined at a given angle, equal to a given triangle.

(3) To construct a parallelogram, with sides inclined at a given angle, equal to a given rectilineal figure.

(4) To construct a square equal to a given rectilineal figure.

PROBLEM I.

To construct a parallelogram equal to a given triangle and having one of its angles equal to a given angle.

Let ABC be the given triangle, E the given angle.

A

B

Construction. Bisect AC in D, make the angle CDF=E, and through B draw BFH parallel to AC, and draw CH parallel to DF

F

FDCH will be the parallelogram required.

E

Proof. If BD be joined, it will be clear that the triangle BAC and the parallelogram FHCD are each of them double of the triangle BDC (II. 2, Cor. 1), and therefore the parallelogram FHCD=the triangle BAC, and it has an angle E, which was required*.

=

* Euclid, 1. 42.

PROBLEM II.

To construct a parallelogram on a given base equal to a given triangle and having one of its angles equal to a given angle.

Let BAC be the given triangle, E the given angle as before, and let it be required to construct on the line GO a parallelogram equal to BAC, and having an angle E.

Construction. Construct the parallelogram FDGH as before, and place it so that one of its sides GH may be in the same straight line with GO.

Produce FD, and draw OQ parallel to GD to meet FD in Q.

4

Join QG, and produce it to meet FH produced in S.

Draw SKR parallel to FQ, meeting DG produced in K, and QO produced in R.

Then GORK is the parallelogram required.

Proof. For the parallelogram FG= the parallelogram GR, being complements (11. 4); and FG=the given triangle ABC.

Therefore GR=the triangle ABC, and it has an angle = E, since it is equiangular with the parallelogram FDGH*.

PROBLEM III.

To construct a parallelogram equal to a given rectilineal figure and having one of its angles equal to a given angle.

* Euclid, I. 44.

Let ABCDE be the given rectilineal figure; F the given angle. Divide ABCDE into triangles by joining BE, BD.

Construction. Construct as before a parallelogram GHIK=BAE, and having an angle at K= F.

Construct on HI a parallelogram HJLI=BED, and having the angle HIL=F.

And construct on JL a parallelogram JMNL = BCD and having the angle JLN= F.

GKNM will then be the parallelogram required.

Proof. For since the angle HIL= the angle K, it is therefore supplementary to HIK; and therefore (by 1. 3) KIL is a straight line.

Similarly GM and KN are straight lines; and MN is parallel to GK.

(I. 24.)

Therefore GKNM is a parallelogram, having the given angle, and it is by construction equal to the given rectilineal figure*.

PROBLEM IV.

To construct a square equal to a given rectilineal figure.

* Euclid, I. 45.