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 (12. 2.), the double of the rectangle contained by DB, BC, has a given ratio to the triangle ABC.

E

Because the angle ABC is given, the angle ABD is also given; and the angle ADB is given; wherefore the triangle ABD is given (43. dat.) in species; and therefore the ratio of AD to DB is given and as AD to DB, so is (1. 6.) 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 A AD, BC: but the ratio of the rectangle AD, BC to the triangle ABC is given, because it is double (41. 1.) of the triangle; therefore the ratio of twice the rectangle DB, BC to the triangle ABC is given (9. dat.); and twice the rectangle DB, BC is the excess (12. 2.) of the square of AC above the squares of AB, BC; therefore this excess has a given ratio to the triangle ABC.

D B

H

F G

C

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 ADB is equiangular to EHF; therefore (4. 6.) as BD to DA, so FH to HE; and as quadruple of BD to DA, so is (cor. 4. 5.) quadruple of FH to HE: but as twice BD is to DA, so is (1. 6.) twice the rectangle DB, BC to the rectangle AD, BC; and as DA to the half of it, so is (cor. 5.) the rectangle AB, BC to its half the

triangle 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.

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.

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

A

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 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 (9. dat.) the ratio of twice the rectangle CB, BD to the triangle ABC is given; and twice the rectangle CB, BD is (13. 2.) 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.

B

D C

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.

A

CASE 1. Bisect the base BC in E, and join AE which will be perpendicular (8. 1.) to BC; wherefore the square of AB is equal (47. 1.) to the squares of AE, EB; but the square of EB is equal (5. 1.) to the rectangle BD, DC together with the square of DE; therefore D B DE

C

the square of AB is equal to the squares of AE, ED, that is, to (47. 1.) the square of AD, together with the rectangle BD, DC; the other case is shown in the same way by 6. 2. Elem.

Ir 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 equal to the rectangle DC, CE together with the square 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 G half (5. & 32.) of the given angle BAC; therefore the triangle ADC is given

D

A

F

C

L

E

K

(43. dat.) in species; and AF is drawn from its vertex to the base in a given angle; wherefore the ratio of AF to the base CD is given (50. dat.); and as CD to AF, so is (1. 6.) the rectangle DC, CE to the rectangle AF, CE; and the ratio of the rectangle AF, CE to its half (41. 1.), the triangle ACE, is given; there fore the ratio of the rectangle DC, CE to the triangle ACE, that is (37. 1.), to the triangle ABC, is given (9. dat.), and the rect angle 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.

The ratio which the rectangle DC, CE has to the triangle ABC is found thus: take the straight line GH given in posi

tion 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 (4. 6. 22. 5.) (DC to AF, and so is) the rectangle DC, CE to the rectangle AF, CE; but as GL 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.

COR. And if a triangle have a given angle, the space 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 the preceding proposition, by help of the second case of the lemma.

If the perpendicular drawn from a given angle of a 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.

If ABC be an isosceles triangle, it is evident (5. & 32. 1.) that

if any one of its angles be given, the rest are also given; and therefore the triangle is given in species, without the consideration of the ratio of the perpendicular to the base, which in this case is given by prop. 50.

But when ABC is not an isosceles triangle, take any straight line EF given in position and magnitude, and upon it describe

* See Note.

the segment of a circle EGF containing an angle equal to the given angle BAC; draw GH bisecting EF at right angles, and join EG, GF: then, since the angle EGF is equal to the angle BAC, and that EGF is an isosceles triangle, and ABC is not, the angle FEG is not equal to the angle CBA: draw EL, making the angle FEL equal to the angle CBA; join FL, and draw LM perpendicular to EF; then, because the triangles ELF, BAC are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB; and as LE to EF, so is AB to BC; wherefore, ex æquali, as LM to EF, so is AD to BC; and because the ratio of AD to BC is given, therefore the ratio of LM to EF is given; and EF is given, wherefore (2. dat.) LM also is given. Complete the parallelogram LMFK; and because LM is given, FK is given in magnitude; it is also given in position, and the point F is given, and consequently (30. dat.) the point K; and because through K the straight line KL is drawn parallel to EF which is given in position, therefore (31. dat.) KL is given in position:

and the circumference ELF is given in position; therefore the point L is given (28. dat.). And because the points L, E, F, are given, the straight lines LE, EF, FL, are given (29. dat.) in magnitude; therefore the triangle LEF is given in species (42. dat.); and the triangle ABC is similar to LEF, wherefore also ABC is given in species.

Because LM is less than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be less than the ratio of GH to EF, which the straight line, in a segment of a circle containing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base.

COR. 1. If two triangles, ABC, LEF, have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the base BC, as the perpendicular LM to the base EF, the triangles ABC, LEF are similar.

Describe the circle EFG about the triangle ELF, and draw LN parallel to EF; join EN, NF, and draw NO perpendicular to EF; because the angles ENF, ELF are equal, and that