If two chords of a circle cut one another, the rectangle contained by the segments of one shall be equal to the rectangle contained by the segments of the other.

Let AB, CD, two chords of the ACBD, cut one another

at E.

Then shall the rect. AE, EB = the rect. CE, ED.
Construction.

Find F, the centre of the ACB; III. 1. From F draw FG, FH perp. respectively to AB, CD. I. 12. Join FA, FE, FD.

Proof.

Because FG is drawn from the centre F perp. to AB,
AB is bisected at G.

III. 3.

For a similar reason CD is bisected at H. Again, because AB is divided equally at G, and unequally at E, ... the rect. AE, EB with the sq. on EG= the sq. on AG. II. 5. To each of these equals add the sq. on GF; then the rect. AE, EB with the sqq. on EG, GF the sum of the sqq. on AG, GF.

But the sqq. on EG, GF = the sq. on FE;
and the sqq. on AG, GF = the sq. on AF;
for the angles at G are rt. angles.

I. 47.

.. the rect. AE, EB with the sq. on FE = the sq. on AF. Similarly it may be shewn that

the rect. CE, ED with the sq. on FE the sq. on FD.

=

But the sq. on AF the sq. on FD; for AF = FD.

.. the rect. AE, EB with the sq. on FE = the rect. CE, ED with the sq. on FE.

From these equals take the sq. on FE:

then the rect. AE, EB the rect. CE, ED. Q.E.D.

=

COROLLARY. If through a fixed point within a circle any number of chords are drawn, the rectangles contained by their segments are all equal.

NOTE. The following special cases of this proposition deserve notice :

(i)

(ii)

(iii)

when the given chords both pass through the centre : when one chord passes through the centre, and cuts the other at right angles:

when one chord passes through the centre, and cuts the other obliquely.

In each of these cases the general proof requires some modification, which may be left as an exercise to the student.

1.

EXERCISES.

Two straight lines AB, CD intersect at Е, so that the rectangle AE, EB is equal to the rectangle CE, ED; shew that the four points A, B, C, D are concyclic.

2. The rectangle contained by the segments of any chord drawn through a given point within a circle is equal to the square on half the shortest chord which may be drawn through that point.

3. ABC is a triangle right-angled at C; and from C a perpendicular CD is drawn to the hypotenuse: shew that the square on CD is equal to the rectangle AD, DB.

4. ABC is a triangle; and AP, BQ, the perpendiculars dropped from A and B on the opposite sides, intersect at O: shew that the rectangle AO, OP is equal to the rectangle BO, OQ.

5. Two circles intersect at A and B, and through any point in AB their common chord two chords are drawn, one in each circle; shew that their four extremities are concyclic.

6. A and B are two points within a circle such that the rectangle contained by the segments of any chord drawn through A is equal to the rectangle contained by the segments of any chord through B: shew that A and B are equidistant from the centre.

7. If through E, a point without a circle, two secants, EAB, ECD are drawn; shew that the rectangle EA, EB is equal to the rectangle EC, ED.

[Proceed as in III. 35, using II. 6.]

8. Through A, a point of intersection of two circles, two straight lines CAE, DAF are drawn, each passing through a centre and terminated by the circumferences: shew that the rectangle CA, AE is equal to the rectangle DA, AF.

If from any point without a circle a tangent and a secant are drawn, then the rectangle contained by the whole secant and the part of it without the circle shall be equal to the square on the tangent.

B

E

F

Let ABC be a circle; and from D, a point without it, let there be drawn the secant DCA, and the tangent DB.

Then the rect. DA, DC shall be equal to the sq. on DB.
Construction. Find E, the centre of the

and from E, draw EF perp. to AD.
Join EB, EC, ED.

ABC:

III. 1.

I. 12.

Proof. Because EF, passing through the centre, is perp. to the chord AC,

.. AC is bisected at F.

III. 3.

And since AC is bisected at F and produced to D, .. the rect. DA, DC with the sq. on FC = the sq. on FD. II. 6. To each of these equals add the sq. on EF:

then the rect. DA, DC with the sqq. on EF, FC = the sqq. on EF, FD. But the sqq. on EF, FC = the sq. on EC; for EFC is a rt. angle; = the sq. on EB.

And the sqq. on EF, FD=the sq. on ED; for EFD is a rt. angle; = the sqq. on EB, BD; for EBD is a

rt. angle.

III. 18.

.. the rect. DA, DC with the sq. on EB = the sqq. on EB, BD. From these equals take the sq. on EB:

then the rect. DA, DC the sq. on DB.

Q.E.D.

NOTE. This proof may easily be adapted to the case when the

secant passes through the centre of the circle.

COROLLARY. If from a given point without a circle any number of secants are drawn, the rectangles contained by the whole secants and the parts of them without the circle are all equal; for each of these rectangles is equal to the square on the tangent drawn from the given point to the circle.

For instance, in the adjoining figure, each of the rectangles PB, PA and PD, PC and PF, PE is equal to the square on the tangent PQ:

.. the rect. PB, PA

=

= the rect. PD, PC

= the rect. PF, PE.

F

B

NOTE. Remembering that the segments into which the chord AB is divided at P, are the lines PA, PB, (see Def., page 139) we are enabled to include the corollaries of Propositions 35 and 36 in a single enunciation.

If any number of chords of a circle are drawn through a given point within or without a circle, the rectangles contained by the segments of the chords are equal.

EXERCISES.

1. Use this proposition to shew that tangents drawn to a circle from an external point are equal.

2. If two circles intersect, tangents drawn to them from any point in their common chord produced are equal.

3. If two circles intersect at A and B, and PQ is a tangent to both circles; shew that AB produced bisects PQ.

4. If P is any point on the straight line AB produced, shew that the tangents drawn from P to all circles which pass through A and B are equal.

5. ABC is a triangle right-angled at C, and from any point P in AC, a perpendicular PQ is drawn to the hypotenuse: shew that the rectangle AC, AP is equal to the rectangle AB, AQ.

V 6. ABC is a triangle right-angled at C, and from C a perpendicular CD is drawn to the hypotenuse: shew that the rect. AB, AD is equal to the square on AC.

If from a point without a circle there are drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained by the whole line which cuts the circle and the part of it without the circle is equal to the square on the line which meets the circle, then the line which meets the circle shall be a tangent to it.

Let ABC be a circle; and from D, a point without it, let there be drawn two st. lines DCA and DB, of which DCA cuts the circle at C and A, and DB meets it; and let the rect. DA, DC the sq. on DB.

Then shall DB be a tangent to the circle.

Construction. From D draw DE to touch the ABC: III. 17. let E be the point of contact.

Find the centre F, and join FB, FD, FE.

III. 1.

Proof. Since DCA is a secant, and DE a tangent to the circle, the rect. DA, DC the sq. on DE, .

=

But, by hypothesis, the rect. DA, DC= the sq. on DB; .. the sq. on DE = the sq. on DB;

III. 36.

Proved. I. Def. 15.

I. 8.

But DEF is a rt. angle, for DE is a tangent; III. 18.

... DBF is also a rt. angle;

and since BF is a radius,

... DB touches the

ABC at the point B.

Q.E.D.