10. PROBLEM. Upon a given straight line, as an hypotenuse, to describe a right-angled triangle, such that the hypotenuse, together with the less of the two remaining sides, shall be the double of the greater of those sides.

Let AB be the given straight line: Upon AB,

as an hypotenuse, it is required to describe a rightangled A, having the less of its two remaining sides, together with AB, the double of the third side.*

Divide (S. 49. 1.) AB into five equal parts in the points C, D, E, F; from A as a centre, at the di

That is, "Upon a given straight line, as an hypotenuse, to describe a right-angled triangle, the sides of which shall be arithmetic proportionals."

stance AF, describe the circle FG, and from B as a centre at the distance BD, describe the circle DG cutting FG in G; join A, G and B, G: The AAGB is right-angled at G, and AB+BG is the double of AG.

For (constr. and S. 7. 2.)
AB AF AE2

= AF+BD (constr.)

= AG + BG (constr. and E. def. 15. 1.) Wherefore (E. 48. 1.) the A AGB is rightangled at G: And since (constr.) AB, and BG, together contain eight of such equal parts as AG contains four, it is manifest that AB+ BG is the double of AG.

PROP. IX.

11. THEOREM. In any triangle, the squares of the two sides are, together, the double of the squares of half the base, and of the straight line joining its bisection and the opposite angle.

Let ABC be any given ▲, of which BC is the

base, and AE the straight line joining the vertex

A, and the bisection E of the base: Then, AB'+
AC2=2AE2+2EB2.

For from A draw (E. 12. 1.) AD to BC, and first let AD fall within the base BC.

Then, BD'+DC=2DE+2EB. (E. 9. 2.)
Add to these equals 2AD'.

... BD2+DC+2AD*=2AD'+2DE*+2EB*•
i. e. AB'+AC2=2AE+2EB. (E. 47. 1.)
And, if the perpendicular AD fall without the
base BC, the proposition may, in like manner, be
deduced from E. 47. 1, and E. 10. 2.

PROP. X.

12. THEOREM. The squares of the sides of any parallelogram are, together, equal to the squares of its diameters taken together.

Let ACBD be a parallelogram, of which AB my

notes on this Prop: are very

learned & her inte

and CD are the diameters: AC+ CB⭑+ BD'+ DA'= AB'+ CD'.

For (S. 42. 1.) AB and CD bisect one another in E:

ACB

.. AC+ CB+ BD'+ DA'= 2AE'+ 2DE'+

2BE*+ 2DE* (S. 9. 2.)

(E. 4. 2, and S. 42. 1.)

AB+ CD'

PROP. XI.

19. THEOREM. If either diameter of a parallelogram be equal to one of the sides about the opposite. angle of the figure, its square shall be less than the square of the other diameter, by twice the square of the other side about that opposite angle.

Let the diameter AB of the ACBD be equal

A

C

D

B

to one of the sides, as AC, about the opposite ZACB; and let CD be the other diameter: Then CD2= AB2+ 2CB3·

For, CD + AB2= 2AC2+ 2CBˆ (S. 10. 2, and E. 34. 1.)

From these equals take AB' which (hyp.) is equal to AC; and there remains,

CD'= AC+2CB':

i.e. CD' AB2+2CB' :

2

Wherefore AB is less than CD' by 2CB'.

PROP. XII.

14. THEOREM. If two sides of a trapezium be parallel to each other, the squares of its diagonals are, together, equal to the aggregate of the squares of its two sides, which are not parallel, and of twice the rectangle of its parallel sides.

Let ABCD be a trapezium, having the side AD

parallel to the side BC, and let AC and BD be its diameters: Then, AC' + BD'= AB' + 2ADXBC.

+ DC2

+

From B and C, the extremities of BC, the greater of the two parallel sides, draw (E. 12. 1.) BE and CF, each 1 tc AD; .. (hyp. and E. 28. 1.) the figure EBCF is a □, and (E. 34. 1.) EF =BC.

First, let both the perpendiculars BE and CF fall without AD, so that both of them meet AD produced.

..

ACDC+AD2+2AD×DF)

and BD'=AB +AD2+2AD× AE

(E. 12. 2.)

.. AC+BD=AB'+DC'+2AD'+2AD XAE

×

+2ADXDF.