PROP. LXXII.

If four straight lines be proportionals; as the first is to the straight line to which the second has a given ratio, so is the third to a straight line to which the fourth has a given ratio.

Let A, B, C, D, be four proportional straight lines, viz. as A to B, so C to D; as A is to the straight line to which B has a given ratio, so is C to a straight line to which D has a given ratio.

Let E be the straight line to which B has a given ratio, and as B to E, so make D to F: The ratio of B to E is given *, and therefore the ratio of D to F; and because as A to B, so is C to D; and as B to E so D to F; therefore, ex æquali, as A to E, so is C to F; and E is the straight line to which B has a given ratio, and F that to which D has a given ratio; therefore as A is to the straight line to which B has a given ratio, so is C to a line to which D has a given ratio.

PROP. LXXIII.

A BE

C D F

If four straight lines be proportionals; as the first is to See N. the straight line to which the second has a given ratio, so is a straight line to which the third has a given ratio to the fourth.

Let the straight line A be to B, as C to D; as A to the straight line to which B has a given ratio, so is a straight line to which C has a given ratio, to D.

Let E be the straight line to which B has a given ratio, and as B to E, so make F to C because the ratio of B to E is given, the ratio of C o F is given: and because A is to B, as to D; and as B to E, so F to C; therefore, ex æquali in proportione perturbata *, A is to E, as F to D; that is, A is to E, to which B has a given ratio, as F, to which C has a given ratio, is to D.

A BE

FCD

23.5.

64.

12.2.

* 43 Dat.

1.6.

41. 1.

Dat.

*12.2.

PROP. LXXIV.

If a triangle have a given obtuse angle; the excess of the square of the side which subtends the obtuse angle, above the squares of the sides which contain it, shall have a given ratio to the triangle.

Let the triangle ABC have a given obtuse angle ABC; and produce the straight line CB, and from the point A draw AD perpendicular to BC: the excess of the square of AC above the squares of AB, BC, that is*, the double of the rectangle contained by DB, BC, has a given ratio to the triangle ABC.

*

Because the angle ABC is given, the angle ABD is also given; and the angle ADB is given; wherefore the triangle ABD is given in species; and therefore the ratio of AD to DB is given: and as AB to DB, so is the rectangle AD, BC, to the rectangle DB, BC; wherefore the ratio of the rectangle AD, BC, to the rectangle DB, BC, is given, as also the ratio of twice the rectangle DB, BC, to the rectangle AD, BC: but the ratio of the rectangle AD, BC, to the triangle ABC, is given, because it is double* of the triangle; there- D B fore the ratio of twice the rectangle

E

FG

DB, BC, to the triangle ABC is given *: and twice the rectangle DB, BC, is the excess of the square of AC above the squares of AB, BC; therefore this excess has a given ratio to the triangle ABC.

And the ratio of this excess to the triangle ABC may be found thus: take a straight line EF given in position and magnitude; and because the angle ABC is given, at the point F of the straight line EF make the angle EFG equal to the angle ABC; produce GF, and draw EH perpendicular to FG; then the ratio of the excess of the square of AC above the squares of AB, BC, to the triangle ABC, is the same with the ratio of quadruple the straight line HF to HE.

Because the angle ABD is equal to the angle EFH and the angle ADB to EHF, each being a right angle; the triangle ABD is equiangular to EHF; therefore * as BD to DA, so FH to HE; and as quadruple of BD *Cor. 4. 5. to DA, so is quadruple of FH to HE: but as twice

* 4. 6.

*

BD is to DA, so is twice the rectangle DB, BC, to 9 Dat. the rectangle AD, BC; and as DA to the half of it, so is the rectangle AD, BC, to its half the triangle * C. 5. ABC; therefore, ex æquali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as quadruple of FH to HE, so is twice the rectangle DB, BC to the triangle ABC.

PROP. LXXV.

If a triangle have a given acute angle, the space by which the square of the side subtending the acute angle is less than the squares of the sides which contain it, shall have a given ratio to the triangle.

65.

Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC, the space by which the square of AC is less than the squares of AB, BC, that is the double of the rectangle contained 13. 2. by CB, BD, has a given ratio to the triangle ABC.

Because the angles ABD, ADB, are each of them given, the triangle ABD is given in species; and therefore the ratio of BD to DA is

DC

given and as BD to DA, so is the rectangle CB, BD to the rectangle CB, AD: therefore the ratio of these rectangles is given, as also the ratio of twice the rectangle CB, BD to the rectangle CB, AD: but the rectangle CB, AD, has a given ratio to its half the triangle ABC: therefore* the ratio of twice the * 9 Dat. rectangle CB, BD, to the triangle ABC is given; and twice the rectangle CB, BD, is the space by which the square of AC is less than the squares of AB, BC; therefore the ratio of this space to the triangle ABC is given and the ratio may be found as in the preceding proposition.

LEMMA.

If from the vertex A of an isosceles triangle ABC, any straight line AD be drawn to the base BC, the square of the side AB is equal to the rectangle BD, DC, of the segments of the base together with the square of AD; but if AD be drawn to the base produced, the square of AD is equal to the rectangle BD, DC, together with the square of AB.

* 13.2.

8.1. 47.1.

5.9

.47.1.

67.

5. and 32. 1.

*

*

*

A or

A

lgonind orto

CASE 1. Bisect the base BC in E, and join AE, which will be perpendicular to BC; wherefore the square of AB is equal to the squares of AE, EB; but the square of EB is equal to the rectangle BD, DC, together with the square of DE; therefore the square of AB is equal to the squares of AE, ED, that is, to the square of AD, together with the rectangle BD, DC. The other case is shewn in the same way by 6. 2. Elem.

PROP. LXXVI.

DBDE

of

CO

If a triangle have a given angle, the excess of the square of the straight line which is equal to the two sides that contain the given angle, above the square of the third side, shall have a given ratio to the triangle

Let the triangle ABC have the given angle BAC, the excess of the square of the straight line which is equal to BA, AC, together above the square of BC, shall have a given ratio to the triangle ABC.

Produce BA, and take AD equal to AC, join DC, and produce it to E, and through the point B draw BE parallel to AC: join AE, and draw AF perpendicular to DC; and because AD is equal to AC, BD is equal to BE; and BC is drawn from the vertex B of the isosceles triangle DBE: therefore by the Lemma, the square of BD, that is, of BA and AC together, is b equal to the rectangle DC, CE, together with the squarer of BC and therefore the square of BA, AC, together, that is, of BD, is greater than the square of BC by the rectangle DC, CE; and this rectangle has a given ratio to the triangle ABC: because the angle BAC is given, the adjacent angle CAD is given; and each of the angles ADC, DCA is given, for each of them is the half* of the given angle BAC; therefore the tri

B

*43 Dat. angle ADC is given in species*; and AF is draw from its vertex to the base in a given angle; wherefore the

ratio of AF to the base CD is given; and as CD to AF, so is the rectangle DC, CE, to the rectangle AF, CE; and the ratio of the rectangle AF, CE, to its half, the triangle ACE, is given; therefore the ratio of the rectangle DC, CE, to the triangle ACE, that is*, to the triangle ABC is given *; and the rectangle DC, CE, is the excess of the square of BA, AC, together, above the square of BC therefore the ratio of this excess to the triangle ABC is given.

* 50 Dat.

1. 6.

41. 1.

*37. 1.

9 Dat.

The ratio which the rectangle DC, CE, has to the triangle ABC is found thus: take the straight line GH given in position and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join KH, and draw GL perpendicular to it: then the ratio of HK to the half of GL is the same with the ratio of the rectangle DC, CE, to the triangle ABC: because the angles HGK, DAC, at the vertices of the isosceles triangles GHK, ADC, are equal to one another, these triangles are similar; and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so is* (DC to AF, and so is) the rectangle DC, CE, to the 22. 5. rectangle AF, CE; but as GE to its half, so is the rectangle AF, CE, to its half, which is the triangle ACE, or the triangle ABC; therefore ex æquali, HK is to the half of the straight line GL, as the rectangle DC, CE, is to the triangle ABC.

space

COR. And if a triangle have a given angle, the by which the square of the straight line, which is the difference of the sides which contain the given angle, is less than the square of the third side, shall have a given ratio to the triangle. This is demonstrated the same way as in the preceding proposition, by help of the second case of the Lemma.

4. 6.

I.

If the perpendicular drawn from a given angle of a See N. triangle to the opposite side, or base, have a given ratio to the base, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC, have a given ratio to it, the triangle ABC is given in species..

D D