PROPOSITION C. THEOREM.

If from any angle of a triangle a straight line be drawn perpendicular to the base, the rectangle, contained by the sides of the triangle, is 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 ▲, and AD the1 from A to BC.
Describe the ABC about the ABC,

draw the diameter AE, and join EC.
Then must rect. BA, AC = rect. EA, AD.
For .* rt. ▲ BDA = ▲ ECA, in a semicircle,
and ABD: = AEC, in the same segment,
..▲ ABD is equiangular to the ▲ AEC.
.. BA is to AD as EA is to AC,
and .. rect. BA, AC=rect. EA, AD.

III. B.

III. 31.

III. 21.

I. 32.

VI. 4.

VI. 16.

Q. E. D.

Ex. 1. Shew that the rectangle contained by the two sides can never be less than twice the triangle.

Ex. 2. ABC is a triangle, and AM the perpendicular upon BC, and P any point in BC; if O, O' be the centres of the circles described about ABP, ACP, the rectangle AP, BC is double of the rectangle of AM, 00'.

Ex. 3. A bisector of an angle of a triangle is produced to meet the circumscribed circle. Prove that the rectangle, contained by this whole line and the part of it within the triangle, is equal to the rectangle contained by the two sides.

PROPOSITION D. THEOREM.

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

B

Let ABCD be any quadrilateral inscribed in a .

Join AC, BD.

Then rect. AC, BD=rect. AB, CD together with rect. AD, BC.

and ▲ BDA =

L

< BCE in the same segment;
..A ABD is equiangular to ▲ BCE,
.. AD is to BD as CE is to BC,
and.. rect. AD, BC=rect. BD, CE.

Again, ABE =

DBC, by construction,

and ▲ BAE

L = ▲ BDC, in the same segment,
..A ABE is equiangular to ▲ BCD.

.. AB is to AE as BD is to CD,

and.. rect. AB, CD=rect. BD, AE.

Hence rect. AB, CD together with rect. AD, BC

=rect. BD, AE together with rect. BD, CE.
=rect. AC, BD.

I. 23.

III. 21.

I. 32.

VI. 4.

VI. 16.

III. 21.

I. 32.

VI. 4.

VI. 16.

II. 1.

Q. E. D.

Ex. If the diagonals cut one another at an angle equal to one third of a right angle, the rectangles contained by the opposite sides are together equal to four times the quadrilateral figure.

PROPOSITION XXVI. THEOREM. (Eucl. vI. 23.)

Equiangular parallelograms have to one another the ratio, which is compounded of the ratios of their sides.

Let AC and CF be equiangulars, having ▲ BCD = 4 ECG.

Then must

AC have to CF the ratio compounded of

the ratios of their sides.

Let BC and CG be placed in a straight line. Then DC and CE are also in a straight line. Complete the

DG, and taking any st. line K,

make as BC is to CG so K to L

and make as DC is to CE so L to M.

I. 14.

VI. 11.

VI. 11.

Then · K has to M the ratio compounded of the ratios of

K to L and L to M,

the sides.

.. K has to M the ratio compounded of the ratios of

VI. Def. 3, p. 260.

PROPOSITION XXVII. THEOREM. (Eucl. VI. 24).

Parallelograms about the diameter of any parallelogram are similar to the whole parallelogram and to one another.

Let ABCD be a, of which the diameter is AC; and AEFG, FHCK the Os about the diameter.

.. AB is to BC as AE is to EF; and since the opposite sides of the Os are equal,

VI. 4.

.. AB is to AD as AE is to AG,

V. 6.

and DC is to CB as GF is to FE,

V. 6.

V. 6.

and CD is to DA as FG is to GA.

Thus the sides of the Os AEFG, ABCD about their equal angles are proportional.

Ex. Show that each of the complements of the parallelogram is a mean proportional between the parallelograms about the diameter.

PROPOSITION XXVIII. THEOREM. (Eucl. VI. 26.)

If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.

Let the

B

s ABCD, AEFG be similar and similarly situated, and have ▲ DAB common.

Then must ABCD and AEFG be about the same diameter.

For, if not, let ABCD have its diameter, AHC, not in the same st. line with AF, the diameter of AEFG.

I. 31.

Let GF meet AHC in H, and draw HK || to AD.
Then s ABCD, AKHG, about the same diameter, are

the less the greater, which is impossible.

.. ABCD and AKHG are not about the same diameter,

and .. ABCD and AEFG must have their diameters in the

same st. line, that is, they are about the same diameter.

Q. E. D.