to the rectangle FE, EG; .. the rectangle FE, EG is equal to the square of ED;

and EF ED :: ED: EG.

(11.) If a chord be drawn parallel to the diameter of a circle, and from any point in the diameter lines be drawn to its extremities; the sum of their squares will be equal to the sum of the squares of the segments of the diameter.

Let CD be drawn parallel to AB the diameter of the circle ACD; and from any point E in AB, let EC, ED be drawn; the squares of EC and ED are together equal to the squares of EA and EB.

F OE G

B

Take O the centre, and join CO, DO; and let fall the perpendiculars CF, DG. Then since CD is parallel to AB, the angles AOC, BOD are equal, and OF=OG. Now (Eucl. ii. 12.) CE CO2 + OE2 + 2 OF × OE,

and (Eucl. ii. 13.) ED2= DO2 + OEˆ − 2 OG × OE, whence the squares of CE, ED are equal to twice the squares of CO, OE, or twice the squares of AO, OE, i. e. to the squares of AE, EB (Eucl. ii. 9.).

(12.) If through a point within or without a circle, two straight lines be drawn at right angles to each other, and meeting the circumference; the squares of the segments of them are together equal to the square of the diameter.

Let AB, CD cut one another at right angles in E; the sum of the squares of AE, EB, CE, ED will be equal to the square of the diameter.

A

E

B

BE

Draw the diameter AF. Join FB, BC, FD, DA; then ABF being a right angle is equal to AED, and .. BF is parallel to CD, and (ii. 1. Cor. 2.) BC=FD. And since the angles at E are right angles, the squares of CE, EB are equal to the square of CB, i. e. to the square of DF; but the squares of AE, ED are equal to the square of AD; .. the squares of CE, EB, AE, ED are equal to the squares AD, DF, i. e. to the square of AF, ADF being a right angle.

(13.) If from a point without a circle there be drawn two straight lines, one of which touches the circle and the other cuts it; and from the point of contact a perpendicular be drawn to the diameter; the square of the line which touches the circle is equal to the square of that part of the cutting line which is intercepted by the perpendicular, together with the rectangle contained by the segments of that part of it which is within the circle.

From the point A without the circle BCD let two lines AB, AC be drawn; of which AB touches the circle, and AC cuts it; and from B let

E

B

D

F

G

BFG be drawn perpendicular to the diameter; the square of AB is equal to the square of AE together with the rectangle CE, ED.

For the square of AB is equal to the squares of AF, FB. But (Eucl. ii. 5.) the square of FB is equal to the rectangle BE, EG together with the square of EF, i. e. to the rectangle CE, ED together with the square of EF;.. the square of AB is equal to the squares of AF, FE together with the rectangle CE, ED, i. e. to the square of AE together with the rectangle CE, ED.

(14.) A straight line drawn from the concourse of two tangents to the concave circumference of a circle is divided harmonically by the convex circumference and the chord which joins the points of contact.

Let AB, AC touch the circle ADC, and AGE cut it. Join BC; then will

AE AG: EF: FG.

On EG as diameter describe a

D

E

H

Б

circle EHG, and through F draw HFI perpendicular to EG. Join AH. Then the rectangle BF, FC is equal to the rectangle EF, FG, i. e. to the square of HF, or the rectangle HF, FI; and .. the are in the circumference of a circle.

points H, B, I, C And since the And since the square

AF, FH, or to the

of AH is equal to the squares of square of AF and the rectangle BF, FC, i. e. to the squares of AK, KF, together with the rectangle BF, FC, i. e. to the squares of AK, KB, or to the square of AB; .. AH=AB; and since the square of AH is equal to the rectangle EA, AG, AH is a tangent at H. And since

EG is a diameter, (ii. 42.)

AE AG: EF: GF.

(15.) If from the extremities of any chord in a circle straight lines be drawn to any point in the circumference meeting a diameter perpendicular to the chord; the rectangle contained by the distances of their points of intersection from the centre is equal to the square described upon the radius.

From A and B, the extremities of the chord AB, let AC, BC be drawn to any point C in the circumference; and let them meet a diameter perpendicular to AB in D and E. Take O

B

FE

the centre; the rectangle DO, OE is equal to the square described on the radius.

Draw the diameter BOG. Join CG, CO. Since the angle OCB is equal to OBC, and BGC to FAD, and that CBG, and BGC together are equal to a right angle; .. OCE and FAD together are equal to a right angle, and .. to FAD and ADF together; hence OCE is equal to ADO, .. the triangles COD, COE are equiangular,

and DO OC :: OC: OE.

.. the rectangle DO, OE is equal to the square of OC.

(16.) If from any point in the base or base produced, of the segment of a circle, a line be drawn making therewith an angle equal to the angle in the segment, and from the extremity of the base any line be drawn to the former, and cutting the circumference; the rectangle contained by this line and the part of it within the segment is always of the same magnitude.

M M

Let ABC be a segment of a circle on the base AC; and from any point D, let DE be drawn, making with AC an angle equal to the angle in the segment, and meeting any line AB drawn from the extremity A; the rectangle EA, AB is of invariable magnitude.

Since the angle ADE is equal to ABC, and the angle at A common to the triangles ADE, ABC, the triangles are.. similar; whence

AD AE AB : AC,

and the rectangle AE, AB is equal to the rectangle AD, AC, which is invariable.

(17.) To determine the locus of the extremities of any number of straight lines drawn from a given point, so that the rectangle contained by each and a segment cut off from each by a line given in position may be equal to a given rectangle.

A

D

B

Let A be the given point, and DE the line given in position. Draw AGF perpendicular to DE; and take AF such that the rectangle AG, AF may be equal to the given rectangle; and on AF as diameter describe a circle; it will be the locus required. Draw any line AC; and join FC. The triangles ABG, AFC being similar,

E

.. the rectangle AC, AB is equal to the rectangle AF, AG, i. e. to the given rectangle. And the same may be proved of any other line drawn from A to the circumference, which.. is the locus.