to a given line in that plane: shew that the straight line joining the feet of the perpendiculars is at right angles to the given line.

10. In how many ways may a solid angle be formed with equilateral triangles and squares?

11. Two planes are inclined to each other at a given angle. Cut them by a third plane, so that its intersections with the given planes shall be perpendicular to each other.

12. AB, AC, AD, are three given straight lines, at right angles to one another. AE is drawn perpendicular to CD, and BE is joined. Shew that BE is perpendicular to CD.

13. Two walls meet at any angle. Shew how to draw on their surfaces the shortest line joining a point on one to a point on the other.

14. Straight lines are drawn from two points to meet each other in a given plane. Find when their sum is the least possible.

15. If two parallel planes be cut by a third plane in the straight lines AB, ab, and by a fourth plane in the straight lines AC, ac respectively, the angle BAC will be equal to the angle bac.

16. If four points be so situated, that the distance between each pair is equal to the distance between the other pair, prove that the angles subtended at any one point by each pair of the others are together equal to two right angles.

17. Give a geometrical construction for drawing a straight line, which shall be equally inclined to three straight lines, meeting at a point.

18. A triangular pyramid stands on an equilateral base. The angles at its vertex are right angles. The square of the perpendicular from the vertex on the base is one-third of the square on either of the edges.

19. If one of the plane angles, forming a solid angle, be a right angle, and the sum of the other two be equal to two right angles, and a plane be drawn, cutting off equal lengths from the two edges, containing the right angle, the sum of the squares on the three straight lines, subtending the plane angles, will be double of the squares on the three edges, containing them.

20. If P be a point in a plane, which meets the containing edges of a solid angle in A, B, C, and O be the angular point, shew that the angles POA, POB, POC are together greater than half the angles AOB, BOC, COA, together.

BOOK XII.

LEMMA.

If from the greater of two unequal magnitudes of the same kind there be taken more than its half, and from the remainder more than its half, and so on, there must at length remain a magnitude less than the smaller of the proposed magnitudes.

Let A and B be two unequal magnitudes of the same kind, of which A is the greater.

Then if from A there be taken more than its half, and from the remainder more than its half, and so on; there must at length remain a magnitude less than B.

Take a multiple of B, as mB, greater than A; and divide A, by the process indicated, taking from it a magnitude greater than its half, and from the remainder a magnitude greater than its half, and carry this process on till there are m divisions, and call the parts successively taken away

C, D, E, F...

Now mB=B, B, B....

.Z

...repeated m times,

and A is greater than the sum of C, D, E,...Z......m in number. Then Z, the last remainder, must be less than B.

For if not, since each of the preceding remainders is greater than Z, each of them would be greater than B, and the sum of C, D.........Z would therefore be greater than mB; that is, A would be greater than mB, which is contrary to the hypothesis. .. Z is less than B.

Q. E. D.

PROPOSITION I. THEOREM.

Similar polygons inscribed in circles are to one another as the squares on the diameters of the circles.

B

M

Let ABCDE, FGHKL be similar polygons inscribed in two Os, and let BM and GN be diameters of the Os.

Then must polygon ABCDE be to polygon FGHKL
as sq. on BM is to sq. on GN.

also, ▲ BAM=▲ GFN, each being a rt. 4,

III. 31.

.. ▲ ABM is equiangular to ▲ FGN,

.. AB is to BM as FG is to GN,

VI. 4.

and .. AB is to FG as BM is to GN.

.. the duplicate ratio of AB to FG-the duplicate ratio

of BM to GN.

VI. 21.

But polygon ABCDE has to polygon FGHKL the duplicate ratio of AB to FG. And sq. on BM has to sq. on GN the duplicate ratio of BM to GN.

..

VI. 21.

· polygon ABCDE is to polygon FGHKL as sq. on BM

is to sq. on GN.

V. 5.

V. 15.

V. 21.

Q. E. D.

PROPOSITION II. THEOREM.

Circles are to one another as the squares on their diameters.

Let ABCD, EFGH be two Os, and BD, FH their diameters:

Then must ○ ABCD be to © EFGH as sq. on BD

is to sq. on FH.

For, if not, sq. on BD must be to sq. on FH as o ABCD is to some space either less than O EFGH, or greater than it. First, if possible, let it be as ABCD is to a space S less than O EFGH.

In

EFGH describe the square EFGH.

This square is greater than half of the EFGH.

IV. 6.

For the sq. EFGH is half of the square, which can be formed by drawing straight lines to touch the circle at the points E, F, G, H; and the square thus formed is greater than the ;

.. sq. EFGH is greater than half of the .