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

ABCD

is to some space either less than ○ EFGH, or greater than it. First, if possible, let it be as ABCD is to a space S less than

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 O;

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

Bisect the arcs EF, FG, GH, HE at the pts. K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE.

Then each of the ▲s EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands.

For ▲ EKF = half of the, formed by drawing a st. line to touch the at K, and parallel st. lines through E and F ; and the thus formed is greater than the segment FEK ; ..▲ EKF is greater than half of the segment FEK, and similarly for the other As.

.. sum of all these triangles is greater than half of the sum of the segments of the O, in which they stand.

Next, bisect EK, KF, etc., and form As as before.

Then the sum of these As is greater than half of the sum of the segments of the O, in which they stand.

If this process be continued, and the As be supposed to be taken away, there will at length remain segments of Os, which are together less than the excess of the O EFGH above the space S, by the Lemma.

Let segments EK, KF, FL, LG, GM, MH, HN, NE be those which remain, and which are together less than the excess of the of the above S.

Then the rest of the O, i.e. the polygon EKFLGMHN, is greater than S.

In

ABCD inscribe the polygon AXBOCPDR similar to the polygon EKFLGMHN.

The polygon AXBOCPDR is to polygon EKFLGMHN as

sq. on BD is to sq. on FH,

XII. 1.

that is, as O ABCD is to the

space S.

Hyp. and V. 5.

But the polygon AXBOCPDR is less than o ABCD,

.. the polygon EKFLGMHN is less than the space S; V.14.

but it is also greater, which is impossible;

.. sq. on BD is not to sq. on FH as ABCD is to any space less than EFGH.

In the same way it may be shown that

sq. on FH is not to sq. on BD as O EFGH is to any space less than ABCD.

Nor is sq. on BD to sq. on FH as ABCD is to any space greater than

EFGH.

B

T

M

For, if possible, let it be as o ABCD is to a space T, greater than EFGH.

Then, inversely, sq. on FH is to sq. on BD as space T is to o ABCD.

But as space T is to o ABCD so is o EFGH to some space, which must be less than ○ ABCD, because space T is greater than EFGH.

V. 14.

.. sq. on FH is to sq. on BD as EFGH is to some space less than ABCD; which has been shewn to be impossible. .. sq. on BD is not to sq. on FH as © ABCD is to any space greater than EFGH.

And it has been shown that

sq. on BD is not to sq. on FH as ○ ABCD is to any space less than o EFGH.

.. sq. on BD is to sq. on FH as o ABCD is to o EFGH.

Q. E. D.

Papers on Euclid (Books VI., XI., and XII.) set in the
Cambridge Mathematical Tripos.

1849. VI. 4. Apply this proposition to prove that the rectangle, contained by the segments of any

1850.

1851.

1852.

chord, passing through a given point within a circle, is constant.

XI. 11. Prove that equal right lines, drawn from a given point to a given plane, are equally inclined to the plane.

VI. 10. AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced, if necessary; if from any point C of AB a perpendicular be drawn to AB, meeting AP and BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional to CE and CF.

VI. 3. If A, B, C be three points in a straight line, and D a point, at which AB and BC subtend equal angles, show that the locus of the point D is a circle.

XI. 8. From a point E draw EC, ED perpendicular to two planes CAB, DAB, which intersect in AB, and from D draw DF perpendicular to the plane CAB, meeting it in F: shew that the line, joining the points C and F, produced if necessary, is perpendicular to AB.

VI. 2. If two triangles be on equal bases, and between the same parallels, any line, parallel to their bases, will cut off equal areas from the two triangles.

1852. XI. 11. ABCD is a regular tetrahedron, and, from

1853.

1854.

1855.

the vertex A, a perpendicular is drawn to the base BCD, meeting it in 0: shew that three times the square on AO is equal to twice the square on AB.

VI. 6. If the vertical angle C, of a triangle ABC, be bisected by a line, which meets the base in D, and is produced to a point E, such that the rectangle, contained by CD and CE, is equal to the rectangle, contained by AC and CB: shew that if the base and vertical angle be given, the position of E is invariable.

XI. 21. If BCD be the common base of two pyramids, whose vertices A and A' lie in a plane passing through BC, and if the two lines AB, AC, be respectively perpendicular to the faces BA'D, CA'D, prove that one of the angles at A, together with the angles at A', make up four right angles.

VL. 16. EA, EA' are diameters of two circles, touching each other externally at E; a chord AB of the former circle, when produced, touches the latter at C', while a chord A'B of the latter touches the former at C: prove that the rectangle, contained by AB and A'B', is four times as great as that contained by BC and B'C.

x1. 20. Within the area of a given triangle is described a triangle, the sides of which are parallel to those of the given one: prove that the sum of the angles, subtended by the sides of the interior triangle, at any point, not in the plane of the triangles, is less than the sum of the angles, subtended at the same point by the sides of the exterior triangle.

vi. 2. A tangent to a circle, at the point A, intersects two parallel tangents in B, C, the points of