THEOREM.*

229. If either angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the sides including the bisected angle, is equal to the square of the bisecting line together with the rectangle contained by the segment of the third side.

Let AD bisect the angle A; then, will

AB.AC AD2+BD.DC.

=

Describe a circle through the three points A, B, C; produce AD till it meets the circumference, and joins CE.

B

D

E

The triangle BAD is similar to the triangle EAC; for by hypothesis, the angle BAD=EAC; also the angle B=E, since they both are measured by half of the arc AC; hence these triangles are similar, and the homologous sides give the proportion, BA : AE :: AD: AC; hence BA.AC-AE.AD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE.AD=AD2 + AD.DE; now AD.DE=BD.DC (224.); hence finally,

BA.AC=AD2 + BD.DC.

THEOREM.

230. In every triangle, the rectangle contained by two sides, is equal to the rectangle contained by the diameter of the circumscribed circle, and the perpendicular let fall upon the third side.

In the triangle ABC, let AD be drawn perpendicular to BC; and let EC be the diameter of the circumscribed circle; then, will

AB.AC=AD.CE.

*This and the three succeeding propositions are not immediately connected with the chain of geometrical investigation. They may be omitted or not, as the reader chooses.-ED.

For, joining AE, the triangles ABD, AEC are right-angled, the one at D, the E other at A; also the angle B=E; these triangles are therefore similar, and they give the proportion, AB : CE :: AD: AC; and hence AB.AC=CE.AD.

231. Cor. If these equal quantities be multiplied by the same quantity BC there will result AB.AC.BC=CE.AD. BC; now AD.BC is double of the surface of the triangle (176.); therefore the product of the three sides of a triangle is equal to its surface multiplied by twice the diameter of the circumscribed circle.

The product of three lines is sometimes called a solid, for a reason that shall be seen afterwards. Its value is easily conceived, by imagining that the lines are reduced into numbers, and multiplying these numbers together.

233. Scholium. It may also be demonstrated, that the surface of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle.

For the triangles AOB,BOC, AOC, which have a common vertex at O, have for their common altitude the radius of the inscribed circle; hence the sum of these triangles will be equal to the sum of the bases AB,BC,

B

E

D

F

AC, multiplied by half the radius OD; hence the surface of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle.

THEOREM.

233. In every quadrilateral inscribed in a circle, the rectangle of the two diagonals is equal to the sum of the rectangles of the opposite sides.

In the quadrilateral ABCD, we shall have

AC.BD=AB.CD+AD.BC.

P

K

B

Take the arc CO=AD, and draw BO meeting the diagonal AC in I..

The angle ABD=CBI, since the one has for its measure half of the arc AD, and the other, half of CO, equal to AD; the angle ADB=BCI, because they are both inscribed in the same segment AOB; hence the triangle ABD is simi

lar to the triangle IBC, and we have the proportion AD: CI :: BD: BC; hence AD.BC=CI.BD. Again, the triangle ABI is similar to the triangle BDC; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc AO=DC; hence the angle ABI is equal to DBC; also the angle BAI to BDC, because they are inscribed in the same segment; hence the triangles ABI, DBC, are similar, and the homologous sides give the proportion, AB : BD:: AI: CD; hence AB.CD-AI.BD.

Adding the two results obtained, and observing that AI.BD +CI.BD=(AI+CI).BD=AC.BD, we shall have AD.BC+

AB.CD=AC.BD.

234. Scholium. Another theorem concerning the inscribed quadrilateral may be demonstrated in the same manner:

The similarity of the triangles ABD and BIC gives the proportion BD: BC: AB: BI; hence BI.BD=BC.AB. If CO be joined, the triangle ICO, similar to ABI, will be similar to BDC, and will give the proportion BD : CO : : DC : OI; hence OI.BD=CO.DC, or because CO=AD,OI.BD= AD.DC. Adding the two results, and observing that BI.BD +OI.BD is the same as BO.BD, we shall have BO.BD= AB.BC+AD.DC.

If BP had been taken equal to AD, and CKP been drawn, a similar train of reasoning would have given us

CP.CA=AB.AD+BC.CD.

But the arc BP being equal to CO, if BC be added to each of them, it will follow that CBP=BCO; the chord CP is therefore equal to the chord BO, and consequently BO.BD and CP.CA are to each other as BD is to CA; hence,

:

BD CA AB.BC+AD.DC AD.AB+BC.CD. Therefore the two diagonals of an inscribed quadrilateral are to each other as the sums of the rectangles under the sides which meet at their extremities.

These two theorems may serve to find the diagonals when the sides are given.

THEOREM.

235. If a point be taken on the radius of a circle, and this radius be then produced, and a second point be taken on it, without the circumference of the circle, these points being so situated, that the radius of the circle shall be a mean proportional between their distances from the centre, then, if lines be drawn from these points to any points of the circumference, the ratio (of these lines) will be constant.

Let P be the point within the circumference, and Q the point without; then if CP: CA:: CA: CQ, the ratio of QM and MP will be the same, for all positions of the point M.

For by hypothesis, CP : CA :: CA: CQ; or substituting CM for CA, CP: CM:: CM: CQ; hence the triangles CPM, CQM, have each an equal angle Q C contained by proportional sides; hence they are similar (208.); and hence the third side

M

M

B

A P C

MP is to the third side MQ, as CP is to CM or CA. But by division, the proportion CP: CA:: CA: CQ gives CP: CA:: CA-CP: CQ-CA, or CP: CA :: AP: AQ; therefore MP MQ :: AP : AQ.

PROBLEMS RELATING TO THE THIRD BOOK.

PROBLEM.

236. To divide a given straight line into any number of equal parts, or into parts proportional to given lines.

First. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefinite straight line AG; and taking AC of any magnitude, apply it five times upon AG; join the last point of division G and the extremity B, by the straight line GB; then draw CI parallel to F GB: AI will be the fifth part of the line AB; G and thus, by applying AI five times upon AB, the line AB will be divided into five equal parts.

E

A

I

M

B

For since CI is parallel to GB, the sides AG, AB, (196.) are cut proportionally in C and I. But AC is the fifth part of AG, hence AI is the fifth part of AB.

P

IF B

Secondly. Let it be proposed to divide the line AB into parts proportional to the given lines P, Q, R. Through A, draw the indefinite line AG; make AC= P, CD=Q, DE=R; join the extremities E and B; and through the points C, D draw CI, DF parallel to EB; the line AB will be divided into parts AI, IF, FB proportional to the given lines P, Q, R.

R

D

ENG

For, by reason of the parallels CI, DF, EB, the parts AI, IF, FB are proportional to the parts AC, CD, DE; and by construction, these are equal to the given lines P, Q, R.