If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

B

D

E

Let ABC be a triangle, and let AD be the perp. from A

to BC:

then the rect. BA, AC shall be equal to the rect. contained by AD and the diameter of the circle circumscribed about the ABC.

ABC;

draw the diameter AE, and join EC.

Describe a circle about the

Then in the AS BAD, EAC,

IV. 5.

the rt. angle BDA = the rt. angle ACE, in the semicircle ACE, ABD = the AEC, in the same segment; III. 21. BAD is equiangular to the EAC;

and the
.. the

I. 32.

VI. 4.

VI. 16.

Q. E.D.

PROPOSITION D. THEOREM.

The rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the two rectangles contained by its opposite sides.

Let ABCD be a quadrilateral inscribed in a circle, and let AC, BD be its diagonals:

then the rect. AC, BD shall be equal to the sum of the rectangles AB, CD and BC, AD.

Make the DAE equal to the BAC;

to each add the

EAC,

I. 23.

and the ABE = the

.. the triangles are equiangular to one another;

.. AB BE :: AC: CD;

.. the rect. AB, CD = the rect. AC, EB.

Again in the ▲ DAE, CAB,

ACD in the same segment; III. 21.

I. 32.

VI. 4.

VI 16.

CAB,

Constr.

III. 21.

I. 32.

VI. 4. VI. 16. Proved.

the

DAE = the

ACB, in the same segment,

.. the triangles are equiangular to one another;

.. the sum of the rects. BC, AD and AB, CD = the sum of

the rects. AG, DE and AC, EB;

that is, the sum of the rects. BC, AD and AB, CD

NOTE. Propositions B, C, and D do not occur in Euclid, but were added by Robert Simson.

Prop. D is usually known as Ptolemy's theorem, and it is the particular case of the following more general theorem:

The rectangle contained by the diagonals of a quadrilateral is less than the sum of the rectangles contained by its opposite sides, unless a circle can be circumscribed about the quadrilateral, in which case it is equal to that sum.

EXERCISES.

1. ABC is an isosceles triangle, and on the base, or base produced, any point X is taken: shew that the circumscribed circles of the triangles ABX, ACX are equal.

2. From the extremities B, C of the base of an isosceles triangle ABC, straight lines are drawn perpendicular to AB, AC respectively, and intersecting at D: shew that the rectangle BC, AD is double of the rectangle AB, DB.

3. If the diagonals of a quadrilateral inscribed in a circle are at right angles, the sum of the rectangles of the opposite sides is double the area of the figure.

4. ABCD is a quadrilateral inscribed in a circle, and the diagonal BD bisects AC: shew that the rectangle AD, AB is equal to the rectangle DC, CB.

5. If the vertex A of a triangle ABC be joined to any point in the base, it will divide the triangle into two triangles such that their circumscribed circles have radii in the ratio of AB to AC.

6. Construct a triangle, having given the base, the vertical angle, and the rectangle contained by the sides.

7. Two triangles of equal area are inscribed in the same circle : shew that the rectangle contained by any two sides of the one is to the rectangle contained by any two sides of the other as the base of the second is to the base of the first.

8. A circle is described round an equilateral triangle, and from any point in the circumference straight lines are drawn to the angular points of the triangle: shew that one of these straight lines is equal to the sum of the other two.

9. ABCD is a quadrilateral inscribed in a circle, and BD bisects the angle ABC: if the points A and C are fixed on the circumference of the circle and B is variable in position, shew that the sum of AB and BC has a constant ratio to BD.

THEOREMS AND EXAMPLES ON BOOK VI.

I. ON HARMONIC SECTION.

1. To divide a given straight line internally and externally so that its segments may be in a given ratio.

H

斤

L MA

Let AB be the given st. line, and L, M two other st. lines which determine the given ratio: it is required to divide AB internally and externally in the ratio L: M.

Through A and B draw any two par1 st. lines AH, BK.

From AH cut off Aa equal to L,

and from BK cut off Bb and Bb' each equal to M, Bʊ' being taken in the same direction as Aa, and Bb in the opposite direction.

Join ab, cutting AB in P;

join ab', and produce it to cut AB externally at Q. Then AB is divided internally at P and externally at Q,

The proof follows at once from Euclid vi. 4.

Obs. The solution is singular; that is, only one internal and one external point can be found that will divide the given straight line into segments which have the given ratio.

DEFINITION.

A finite straight line is said to be cut harmonically when it is divided internally and externally into segments which have the same ratio.

A

Thus AB is divided harmonically at P and Q, if

AP PB AQ: QB.

P and Q are said to be harmonic conjugates of A and B.

If P and Q divide AB internally and externally in the same ratio, it is easy to shew that A and B divide PQ internally and externally in the same ratio: hence A and B are harmonic conjugates of P and Q.

Example. The base of a triangle is divided harmonically by the internal and external bisectors of the vertical angle:

for in each case the segments of the base are in the ratio of the other sides of the triangle. [Euclid vi. 3 and A.]

Obs. We shall use the terms Arithmetic, Geometric, and Harmonic Means in their ordinary Algebraical sense.

1. If AB is divided internally at P and externally at Q in the same ratio, then AB is the harmonic mean between AQ and AP.

2. If AB is divided harmonically at P and Q, and ○ is the middle point of AB;

then shall OP. OQ=OAo.

or,

B

For since AB is divided harmonically at P and Q,

.. AP: PB AQ: QB;

.. AP - PB: AP+PB=AQ - QB : AQ+QB,

Conversely, if

it may be shewn that

2OP: 20A 20A : 20Q;

.. OP. OQ=OA2.

OP. OQ=OA2,

AP: PB AQ: QB;

that is, that AB is divided harmonically at P and Q.