Again, because

AD+ DB2+2 AD × DB=AB2)

AC2+CB2+2 AC × CB = AB2)

(E. iv. 2.)

and that, as has been shewn AD × DB < AC × CB. .. AD2+DB2 > AC2 + CB2.

Lastly, since

AD2+DB2+2 AD × DB

=AC2+CB+2AC × CB,

it is manifest, if from these equals there be taken AC2 + CB2+2 AD × DB, that the excess of AD +DB2 above AC2+ CB2 is the double of the excess of AC CB above AD × DB.

PROP. III.

4. THEOREM. In any isosceles triangle, if a straight line be drawn from the vertex to any point in the base, the square upon this line, together with the rectangle contained by the segments of the base, is equal to the square upon either of the equal sides. Let ABC be an isosceles triangle, and let AQ, be

drawn from its vertex A, to any point Q, in BC its base: AQ2+BQ × QC= AB2.

For bisect (E. x. 1.) BC in D, and join A, D. ··. QD2 + BQ × QC= BD (constr. and E. v. 2.)

To each of these equals add DA;

.. AD2 + QD2+BQ × QC= AD2+DB2: ' But (constr.) BD=DC, and DA is common to the as ADB, ADC, and (hyp.) AB= AC; therefore the ADB ADC; and therefore each of these angles is a right angle; therefore (E. xlvii. 1.) AD2 +DQ2 = AQ2, and AD + DB2= AB2 ;

.. AQ2+BQ × QC=AB2.

PROP. IV.

5. THEOREM. The rectangle contained by the aggregate and the difference of two unequal straight lines is equal to the difference of their squares.

Let AC and CB be two unequal straight lines,

of which CB is the greater; and let them be placed in the same straight line AB; so that AB is the aggregate of AC, CB, and if (E. iii. 1.) CD be cut off from CB equal AC, DB is the difference between AC and CB. Then since (constr. and E. vi. 2.)

AB × DB+AC2 = CB2, it is manifest, if from these equals AC be taken, that AB DB = CB2 - AC2;

i. e. the rectangle contained by the aggregate AB, of AC and CB, and their difference DB, is equal to the difference of their squares.

6. COR. If there be three straight lines, the difference between the first and second of which is equal to the difference between the second and third, the rectangle contained by the first and third, is less

than the square of the second, by the square of the common difference between the lines.

For, let AB, CB, and DB be the three straight lines, having AC the difference of AB and CB, equal to CD, the difference of CB and DB: Then, since it has been shewn that AB × DB= CB3 ~ AC2, it is manifest that AB × DB is less than CB by AC2.

PROP. V.

7. THEOREM. The square of the excess of the greater of two given straight lines above the less, is less than the squares of the two lines, by twice the rectangle contained by them.

For let AB and CB be two straight lines, of

which AB is the greater: Then is AC the excess of AB above CB; and since (E. vii. 2.) AC2 +2 AB × BC= AB2 + CB, it is manifest that AC is less than AB2 + CB2 by 2AB × BC.

PROP. VI.

8. THEOREM. The squares of any two unequal straight lines are, together, greater than twice the rectangle contained by those lines.

For let AB and CB be two straight lines of

which AB is the greater: Then since (E. vii. 2.) AB + CB2 = 2AB × BC+AC2

it is manifest that AB+ CB > 2AB × BC.

PROP. VII.

9. THEOREM. If a straight line be divided into five equal parts, the square of the whole line is equal to the square of the straight line, which is made up of four of those parts, together with the square of the straight line which is made up of three of those parts.

Let the straight line AB be divided into five equal

parts by the points C, D, E, F: Then, AF+AЕ2

= AB3.

For since (hyp.) EF=FB;

.. 4 FE × AF÷AE2 = AB2 (E. viii. 2.) But, since AC=CD=DE=EF, 4FE=AF; .. 4FE × AF=AF2;

.. AF+AE" = AB°.

PROP. VIII.

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 triangle, having the less of its two remaining sides, together with AB, the double of the third side*.

Divide (Supp. xlix. 1.) AB into five equal parts in the points C, D, E, F; from A as a centre, at the distance AF, describe the circle FG, and from Bas a centre at the distance BD, describe the circle DG cutting FG in G; join A, G and B, G: The ▲ AGB is right-angled at G, and AB+BG is the double of AG.

For (constr. and Supp. vii. 2.)

ABAF+AE2

=AF+BD (constr.)

= AG2+ BG (constr. and E. def. 15. 1.) Wherefore (E. xlviii. 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 triangle, of which BC is the base, and AE the straight line joining the vertex A, and the bisection E of the base: Then,

AB2 + AC2 = 2 AE2+2 EB2.

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