Imágenes de páginas
PDF
EPUB

of them, if the side opposite the given angle is shorter than the other given side, two different triangles may be formed, each of which will have the given parts. This is called the ambiguous case.

Suppose that the side a and side c, and X C, are given. If a>c, then, making the * C, and cutting off CB = a, taking B

[merged small][merged small][merged small][ocr errors][merged small]

as center, and describing an arc with radius equal to c, it may cut CD in two points; and the two unequal triangles ABC and A'BC will satisfy the required conditions.

In these the angles opposite the side BC are supplemental, by 177

Loci.

181. The aggregate of all points and only those points which satisfy a given condition, is called the Locus of points satisfying that condition.

182. Hence, in order that an aggregate of points, L, may be properly termed the locus of a point satisfying an assigned condition, C, it is necessary and sufficient to demonstrate the following pair of inverses :

(1) If a point satisfies C, it is upon L. (2) If a point is upon L, it satisfies C.

We know from 18, that, instead of (1), we may prove its contranominal :

(3) If a point is not upon L, it does not satisfy C.

Also, that, instead of (2), we may prove the obverse of (1):

(4) If a point does not satisfy C, it is not upon L.

THEOREM XXV.

183. The locus of the point to which sects from two given points are equal is the perpendicular bisector of the sect joining them.

[ocr errors][ocr errors][ocr errors][merged small][merged small]

CONCLUSION. PC I AB.
PROOF. Join PC.
Then A ACP is = ABCP,

(129. Triangles with the three sides respectively equal are congruent.)

..

:: ACP = 4 BCP,

CP is the perpendicular bisector of AB.
184. INVERSE.
HYPOTHESIS. P, any point on CPI AB at its mid-point C.
CONCLUSION. PA PB.

PROOF. A ACP ABCP. (124. Two triangles are congruent if two sides and the included angle are equal in

each.)

THEOREM XXVI.

185. The locus of points from which perpendiculars on two given intersecting lines are equal, consists of the two bisectors of the angles between the given lines.

[ocr errors][merged small][merged small][merged small][merged small]

HYPOTHESIS. Given AB and CD intersecting at O, and P any point such that PM I AB = PN I CD.

CONCLUSION. POM * PON.
PROOF. Δ ΡΟΜ Δ Δ ΡΟΝ.

(179. Right triangles having hypothenuse and one side equal are congruent.)

Therefore all points from which perpendiculars on two intersecting lines are equal, lie on the bisectors of the angles between them.

186. INVERSE.

HYPOTHESIS. P, any point on bisector.
CONCLUSION. PM

= PN. PROOF. A POM - PON. (176. Triangles having two angles and a corresponding side equal in each are

congruent.)

Hence, by 110, the two bisectors of the four angles between AB and CD are the locus of a point from which perpendiculars on AB and CD are equal.

Intersection of Loci.

187. Where it is required to find points satisfying two conditions, if we leave out one condition, we may find a locus of points satisfying the other condition.

Thus, for each condition we may construct the corresponding locus; and, if these two loci have points in common, these points, and these only, satisfy both conditions.

THEOREM XXVII.

188. There is one, and only one, point from which sects to three given points not on a line are equal.

[ocr errors][ocr errors][merged small][merged small][ocr errors][merged small]

HYPOTHESIS. Given A, B, and C, not in a line.

CONSTRUCTION AND PROOF. By 183, every point to which sects from A and B are equal, lies on DD, the perpendicular bisector of AB. Again: the locus of points to which sects from B and C are equal is the perpendicular bisector FF'.

Now, DD' and FF' intersect, since, if they were parallel, AB, which is perpendicular to one of them, would be perpendicular to the other also, by 170, and ABC would be one line. Let them intersect in O. By 95, they cannot intersect again. 189. Thus,

OB

OA, and

OC =

OA,

[blocks in formation]

Therefore, by 183, O is on the perpendicular bisector of AC. Therefore

COROLLARY. The three perpendicular bisectors of the sides of a triangle meet in a point from which sects to the three vertices of the triangle are equal.

PROBLEM X.

190. To find points from which perpendiculars on three given lines which form a triangle are equal.

[ocr errors][ocr errors][ocr errors]

GIVEN, a, b, c, three lines intersecting in the three distinct points A, B, C.

SOLUTION. The locus of points from which perpendiculars on the two lines a and b are equal consists of the two bisectors of the angles between the lines.

The locus of points from which perpendiculars on the two lines b and c are equal, similarly consists of the two lines bisecting the angles between 6 and c. Our two loci consist thus of two pairs of lines. Each line of one pair cuts each line of the second pair in one point; so that we get four points, 0, 01, 02, 03, common to the two loci.

191. By 185, the four intersection points of the bisectors of the angles between a and b, and between b and c, lie on the bisectors of the angles between a and c. Therefore COROLLARY. The six bisectors of the interior and exterior

. angles of a triangle meet four times, by threes, in a point.

« AnteriorContinuar »