and therefore bisect one another; hence m = = BE and

EO, and nO DO; therefore E O

=

If, now, we take a similar construction with DC and the third middle line, A F, we shall be able to prove that A F cuts DC in a point one-third of its length from D; therefore the third middle line also passes through the point O. Hence the three middle lines intersect in the same point which divides each into two parts, one of which is double of the other.

63. If the opposite sides of any quadrilateral be produced to meet, what is the condition that the bisectors of the two angles formed shall be at right angles to one another?

Let A, B, C, D be the angles of the quadrilateral; A being the greatest, and E and F the angles formed by the produced sides. Let the bisectors of E and F intersect at right angles at O; then, from the fact that the exterior angle of a triangle is equal to the sum of the interior and opposite angles, if R be written for a right angle,

Also, by taking exterior angles at O,

2

a right angle = ¦ E + ¦ F + (2 R – A), or 2 A 2 R = E + F,

=

2 A+B+ D-4R;

.. B+ D = 2 R,

and A+ C: = 2 R,

the required condition.

64. Determine the path of a ray of light that, after reflection at two mirrors, it may return to the same point.

Let A B, AC be the right sections of the mirrors, and P the point. Draw PB and PC perpendicular to AB and A C, and continue PB to Q, making BQ=B P, and PC to R, making C R = C P. Join QR, and let it cut A B in D and AC in E, and then join P D, D E, E P.

It will readily be proved that PD, DE make equal angles with A B, and D E, E P make equal angles with A C. Hence PDEP is the angle required.

3. Through a given point draw a straight line which shall make equal angles with two intersecting lines.

Let AB, AC be the straight lines and G the given point.

Draw A M bisecting the angle BA C, and through G draw EF perpendicular to AM. The triangles AEM, AFM are equal and AEM = AFM.

с

M

4. Find a point in a given straight line at the same distance from each of two other straight lines.

E

Let E F be the given straight line, and let A be the point of intersection of the other straight lines. Let AD be the bisector of the angle A, and suppose A D to meet the straight line E F in the point B; B is the point required. Every point which

A

B

is equidistant from the two straight lines meeting at A lies in the line A D by the previous question; but B is the only point in the line E F which is also in the line A D; therefore B is the only point in EF which is equidistant from the two given straight lines.

CHAPTER VIII.

PROBLEMS.

1. Through a given point to draw a straight line which shall make a given angle with a given straight line.

Let A be the given point and BC the given straight line.

B F

D

E Take any point, D in BC, and make CDE equal to the given angle. Through A draw A F parallel to E D.

Then it is evident that A FD= EDC, for they are corresponding angles; and, EDC being equal to the given angle, .. A FD is equal to the given angle.

2. Find a point in a line which shall be equidistant from two given points.

Let A, B be the points and M the given line. At C, the middle point of A B, draw CD perpendicular to A B, and meeting the line M in D; then AD=DB.