B

sides in each, the triangles are not necessarily con

A P

A'

1. If BC is > AB,

ДА'В'С' = ∆АВС.

Proof. Since BC is > AB, therefore B'C' is > A'B'. Place A' on A and A'C' along AC.

LA'=LA, and A'B'=AB,

Let

B' coincides with B.

BP be 1 AC;

(34°, 27°)

then B'C' cannot lie between BA and BP (63°, 2), but must lie on the same side as BC; and being equal to BC, the lines B'C' and BC coincide (63°, 3), and hence

Now, let A'B'C' be superimposed on AABC so that A' coincides with A, B' with B, and A'C' lies along AC. Then, since we are not given the length of A'C', B'C' may coincide with BC, and the As A'B'C' and ABC be congruent; or B'C' may coincide with BD, and the triangles A'B'C' and ABC be not congruent.

q.e.d.

Hence when two triangles have two sides in the one respectively equal to two sides in the other, and an angle opposite one of the equal sides in each equal, the triangles are not necessarily congruent unless some other relation exists between them.

The first part of the theorem gives one of the sufficient relations. Others are given in the following corollaries.

Cor. 1. If LC is a 7, BC and BD (2nd Fig.) coincide along BP, and the As ABD and ABC become one and the same. Hence C'must fall at C, and the As A'B'C' and ABC are congruent.

Cor. 2. The LBDA is supplementary to BDC and therefore to BCA. And. LBDA is > LBPA, .. LBDA is greater than a right angle, and the BCA is less than a right angle.

Hence if, in addition to the equalities of the theorem, the angles C and C' are both equal to, or both greater or both less than a right angle, the triangles are congruent.

Def.-Angles which are both greater than, or both equal to, or both less than a right angle are said to be of the same affection.

66°. A triangle consists of six parts, three sides and three angles. When two triangles are congruent all the parts in the one are respectively equal to the corresponding parts in the other. But in order to establish the congruence of two triangles it is not necessary to establish independently the respective equality of all the parts; for, as has now been shown, if certain of the corresponding parts be equal the equality of the remaining parts and hence the congruence of the triangles follow as a consequence. Thus it is sufficient that two sides and the included angle in one triangle shall be respectively equal to two sides and the included angle in another. For, if we are given these parts, we are given consequentially all the parts of a triangle, since every triangle having two sides and the included angle equal respectively to those given is congruent with the given triangle.

Hence a triangle is given when two of its sides and the angle between them are given.

A triangle is given or determined by its elements being given according to the following table :

1. Three sides,

2. Two sides and the included angle,

(58)

(52°)

(59°)

(64°)

3. Two angles and the included side, 4. Two angles and an opposite side, 5. Two sides and the angle opposite the longer side, (65°)

•

When the three parts given are two sides and the angle opposite the shorter side, two triangles satisfy the conditions, whereof one has the angle opposite the longer side supplementary to the corresponding angle in the other.

This is known as the ambiguous case in the solution of triangles.

A study of the preceding table shows that a triangle is completely given when any three of its six parts are given, with two exceptions:

(1) The three angles;

(2) Two sides and the angle opposite the shorter of the two sides.

67°. Theorem.-If two triangles have two sides in the one respectively equal to two sides in the other, but the included angles and the third sides unequal, then

I. The one having the greater included angle has the greater third side.

2. Conversely, the one having the greater third side has the greater included angle.

Let BE bisect the DBC and meet AC in E.

2. AC is > A'C', then LABC is greater than LA'B'C'. Proof.--The proof of this follows from the Rule of Identity.

68°. Theorem.—1. Every point upon a bisector of an angle is equidistant from the arms of the angle.

2. Conversely, every point equidistant from the arms of an angle is on one of the bisectors of the angle.

1. OP and OQ are bisectors of the angle AOB, and PA, PB are perpendiculars from P upon

B

P

A

If Q be a point on the bisector OQ it is shown in a similar manner that the perpendiculars from Q upon the arms of the angle AOB are equal.

q.e.d.

2. If PA is to OA and PB is 1 to OB, and PA=PB, then PO is a bisector of the angle AOB.

Proof. The As POA and POB are congruent, since they have two sides and an angle opposite the longer equal in each (65°, 1); .. ¿POA=¿POB,

and PO bisects the LAOB,

Similarly, if the perpendiculars from Q upon OA and OB are equal, QO bisects the BOA', or is the external bisector of the LAOB.

q.e.d.

LOCUS.

69°. A locus is the figure traced by a variable point, which takes all possible positions subject to some constraining condition.

If the point is confined to the plane the locus is one or more lines, or some form of curve.

Illustration. In the practical process of drawing a line or curve by a pencil, the point of the pencil becomes a variable (physical) point, and the line or curve traced is its locus.

In geometric applications the point, known as the generating point, moves according to some law.

The expression of this law in the Symbols of Algebra is known as the equation to the locus.

Cor. 1. The locus of a point in the plane, equidistant from the end-points of a given line-segment, is the right bisector of that segment.

This appears from 54°.

Cor. 2. The locus of a point in the plane, equidistant from two given lines, is the two bisectors of the angle formed by the lines.

This appears from 68°, converse.

EXERCISES.

1. How many lines at most are determined by 5 points? by 6 points? by 12 points?

2. How many points at most are determined by 6 lines? by

12 lines?

3. How many points are determined by 6 lines, three of which

pass through a common point?