From A draw a line AC equal and parallel to MN. Then AC is the distance that must be travelled in the given direction; the remaining distance is to be as short as possible, and so will be equal to the linesegment CB.

Let CB cut q at E. Draw ED parallel to NM or CA, meeting p at D. Join AD. Then ADEB is the required path.

The pupil can easily prove that ACED is a parallelogram, and hence that the sum of the distances AD, DE, and EB is equal to the sum of the distances AC and CB. The distance travelled in going from A to B would have been the same if we had crossed from the line AD to the parallel line CB, in the given direction, at any point, and the only problem was to shift the given distance AC parallel to itself till it reached from one bank of the river to the other.

When a straight line is moved so as to remain always parallel to its original position it is said to be translated, and the motion is called a translation.

When any figure is moved so that all of its points describe parallel straight lines it is said to be translated.

Since the solution of this problem required the translation of the line AC, which satisfied some of the conditions of the problem into such a position as to satisfy the remaining conditions, the method is sometimes called the method of Translation.

EXERCISES

1. Through a given point to draw a straight line to make equal angles with two given intersecting lines.

SUGGESTION.

Translate the bisector of the angle between the lines till it passes through the given point. See another solution on p. 66.

2. Construct a triangle having given one side and the orthocentre.

3. Construct a triangle having given the base, the altitude, and the length of one side. Is the solution always possible?

4. Through a given point P draw a straight line such that the perpendiculars to it from two fixed points meet it at equal distances from P.

5. Through a given point draw a line-segment terminated by two given intersecting straight lines, which shall be bisected at the given point.

6. Construct a right triangle having given the hypotenuse and the difference of the two sides.

7. Construct a triangle equiangular with a given triangle and having a given perimeter.

SECTION VI

SYMMETRICAL FIGURES

144. If there be given any straight line a, and from a point A we draw AM perpendicular to the given line, and produce it to a point A' such that MA' equals MA, then A and A' are called inverse points relative to the

145. If you think of the plane in which there lie two figures symmetrical with respect to a certain straight line, as being folded along this line, just as you would fold a sheet of paper along a given straight line, it is clear that the two figures would come to coincide, since all lines perpendicular to the axis of symmetry would fold over upon themselves.

Hence,

THEOREM. If two figures are symmetrical with respect to a straight line, they are superposable by inversion.

A figure symmetrical with a given figure can easily be obtained by tracing it with ink and then folding the paper along a straight line before the ink is dry.

146. Two parts of the same figure may be symmetrical with respect to a given straight line, as in the diagram. In such cases we say that the figure is symmetrical with respect to the line.

THEOREM.

A circle is symmetrical with respect

to any of its diameters.

Let APB be a circle whose centre is 0, AB

any diameter, P any point on the circle.

From P draw PM perpendicular to the diame

ter, and produce it to meet the circle again at P'. Join PO and P'O.

The As OPM and OP'M are identically equal. Hence P' is the inverse point of P, and for every point of the circle there is an inverse point with respect to the diameter.

EXERCISES

1. Show that a square is symmetrical with respect to each of its diagonals.

2. Show that an equiangular triangle is symmetrical with respect to each of its medians.

3. If a quadrilateral is symmetrical with respect to one of its diagonals, show that the two diagonals are at right angles.

4. If a triangle is symmetrical with respect to a median, it must be isosceles.

MISCELLANEOUS EXERCISES

1. In a quadrilateral ABCD, if AB is the greatest side and CD the least, show that BCD is greater than DAB, and ≤ CDA greater than

ZABC.

2. Show that the sum of the diagonals of a convex quadrilateral is greater than the sum of either pair of opposite sides, and also greater than half the sum of the four sides.

3. The sum of the distances of any point within a rectilinear figure from the vertices is greater than half the sum of the sides.

4. If D is any point on the side BC of a triangle ABC, show that the sum of the sides of the triangle is greater than twice AD.

5. Show that no convex polygon can have more than three of its interior angles acute, nor more than three of its exterior angles obtuse.

6. On a given straight line find a point which is equidistant from two given lines. Under what conditions is the solution impossible?

7. Show how to find a point which is at a given distance from each of two intersecting straight lines. How many such points are there?

8. If two quadrilaterals are mutually equiangular, and two adjacent sides of the one are respectively equal to the two corresponding adjacent sides of the other, show that the quadrilaterals are identically equal.

9. Show that the sum of any two medians of a triangle is greater than the third median.

10. Show that the sum of the three medians of a triangle is greater than three-fourths of the sum of the sides of the triangle.

11. Show that if two of the medians of a triangle are equal, the triangle must be isosceles.

12. If E and F are the mid-points of the sides AB and CD of the parallelogram ABCD, show that the lines ED and BF will trisect the diagonal AC.

13. If the vertices of one parallelogram lie on the sides of another, show that all four diagonals pass through the same point.

14. If an exterior angle of a triangle be bisected, and also one of the interior non-adjacent angles, the angle made by the two bisectors is equal to half of the other interior non-adjacent angle.

15. If two triangles on the same side of a common base have their sides which are terminated in opposite extremities of the base equal, the line joining the vertices will be parallel to the common base.

16. From the vertex A of any triangle ABC two straight lines are drawn meeting the opposite side at D and E, so as to make the angle BAD equal to the angle C, and the angle CAE equal to the angle B. Show that the triangle DAE is isosceles.

17. In any triangle ABC, the bisector of the angle A makes with the perpendicular drawn from A to the opposite side an angle equal to half the difference between the angles B and C.

18. In any triangle ABC, the bisectors of the angles B and C make with each other an angle greater by a right angle than one-half A.

19. In any triangle ABC, the bisectors of the exterior angles at B and C make with each other an angle less by a right angle than one-half A. 20. In any convex quadrilateral, the bisectors of two consecutive angles make with each other an angle equal to the half-sum of the other two angles.

21. If through the point of intersection of the bisectors of the angles B and in any triangle ABC, a line-segment MN is drawn parallel to the side BC, and terminated by the sides AB and AC, show that MN equals the sum of BM and CN.

22. If upon one boundary OX of an angle you choose two points A and A', and upon the other boundary OY you choose two points B and B', such that OB equals OA and OB' equals OA', and join crosswise AB' and A'B, show that the intersection of these lines is upon the bisector of the angle ΧΟΥ.

23. If two sides of a triangle are unequal, the median through their intersection makes the greater angle with the lesser side.

24. A quadrilateral which has two sides equal and the other two sides parallel may be a parallelogram, or it may be a trapezoid having an axis of symmetry.

25. In an isosceles triangle the sum of the distances of a point in the base from the two sides is the same, no matter where in the base the point is chosen.

If the point should be chosen in the base produced, how would this theorem be altered?

26. In an equilateral triangle the sum of the distances of a point within the triangle from the three sides is the same, no matter where the point is chosen.