35.

Second Test for Congruence of Triangles.

If two triangles have two angles and the included side of one equal respectively to two angles and the included side of the other, the triangles are congruent. This is shown by the following argument:

Let ABC and A'B'C' be two triangles in which ▲ A=ZA', B=B', and AB= A'B'.

We are to show that ▲ ABC≈ ▲ A'B'C'.

Place A ABC upon ▲ A'B'C' so that AB coincides with its equal A'B', making C fall on the same side of A'B' as C'. Then AC will take the direction of A'c', since A = A', and the point C must fall somewhere on the ray A'c'.

Also BC will take the direction of B'C' (Why?), and hence C must lie on the ray B'C'.

Since the point c lies on both of the rays A'C' and B'C', it must lie at their point of intersection d' (§ 5). Hence, the triangles coincide and are, therefore, congruent (§ 27).

1. In the figure of § 35 is it necessary to move AABC out of the plane in which the triangles lie? Is it necessary in the figure here given?

2. Show how to measure the height of a tree by using the second test for congruence.

SUGGESTION. Lay out a triangle on the ground which is congruent to ▲ ABC', using § 35.

3. By the second test determine whether A OHGA OJK on page 4.

4. Draw any triangle. Construct another tri

Β'

B

B

angle congruent to it. Use § 35 and also § 32. Use the protractor to construct the angles.

5. Find the distance A C, when C is inaccessible.

Let B be a convenient point from which A and C are visible. Lay out a triangle ABC' making 23 21 and 4 = 22. Show that the distance AC may be found by measuring AC'.

=

6. Show how to find the distance between two inaccessible points A and B.

SOLUTION. Suppose that both A and B are visible from C and D. (1) Using the triangle CDA, find the length of AD as in Ex. 5 above. (2) Using the triangle CBD, find DB in the same manner. (3) Using the triangle DBA, find AB as in Ex. 5, § 34.

37. The proof of the third test for congruence of triangles involves the following:

B

2

B

The angles opposite the equal sides of an isosceles triangle are equal.

Let ABC be an isosceles triangle having

The theorems § 35 and § 37 are due to Thales.

B

It is said he used

§ 35 in calculating the distance from the shore to a ship at sea.

38.

On page 4 pick out as many be shown to be equal by § 37.

EXERCISE.

pairs of angles as possible which may Test these by using the protractor.

39. Definitions. Two angles which have a common vertex and a common side are said to be adjacent if neither angle lies within the other.

Thus, 1 and 2 are adjacent, while 21 and 23 are not adjacent. The sum of two angles is the angle formed by the sides. not common when the two angles are placed adjacent. Thus, 4321+ 22.

If 23 = 21 + either 1 or 2.

2, then we say that 23 is greater than This is written 23 > 1 and 3> <2. An angle may also be subtracted from a greater or equal angle. Thus if ≤3 = 21+ ≤2, then 3-1 = ≤2 and ≤3 − ≤2 = 21. It is clear that:

If equal angles are added to equal angles, the sums are equal angles.

Angles may be multiplied or divided by a positive integer as in the case of line-segments. See § 10.

40. We may now prove the third test for congruence of triangles, namely:

If two triangles have three sides of one equal respectively to three sides of the other, the triangles are congruent.

Let ABC and A'B'C' be two triangles in which AB = A'B', BC= B'C', CA = C'A'.

We are to show that A ABC ≈ ▲ A'B'C'.

Place A A'B'C' so that A'B' coincides with AB and so that c' falls on the side of AB which is opposite c.

(Why is it possible to make A'B' coincide with AB?)

Draw the segment cc'. From the data given, how can § 37 be used to show that in ▲ ACC' ≤1 = <2?

Use the same argument to show that 23 24.

21=22

=

But if

and

23 = 24,

Make an outline of the steps in the above argument, and see that each step is needed in deriving the next.

41. Definition. If one triangle is congruent to another because certain parts of one are equal to the corresponding parts of the other, then these parts are said to determine the triangle. That is, any other triangle constructed with these given parts will be congruent to the given triangle.

1. In § 37 show that CD is perpendicular to AB and that AD=DB. State this fully in words.

2. Using § 40, determine which of the following triangles on page 4 are congruent: OJK, HNP, OIH, PHG, JKU.

3. Do two sides determine a triangle? Three sides? Two angles? Three angles? Illustrate by figures.

4. A segment drawn from the vertex of an isosceles triangle to the middle point of the base bisects the vertex angle and is perpendicular to the base.

5. What parts of a triangle have been found sufficient to determine it? In each case how many parts are needed?

43. The three tests for congruence of triangles, §§ 32, 35, 40, lie at the foundation of the mathematics used in land surveying. The fact that certain parts of a triangle determine it shows that it may be possible to compute the other parts when these parts are known. Rules for doing this are found in Chapter III.

CONSTRUCTION OF GEOMETRIC FIGURES.

44. The straight-edge ruler and the compasses are the instruments most commonly used in the construction of geometric figures.

By means of the ruler straight lines are drawn, and the compasses are used in laying off equal line-segments and also in constructing arcs of circles (§ 12).

Other common instruments are the protractor (§ 33) and the triangular ruler with one square corner or right angle. The three tests for congruence of two triangles are of constant use in geometrical constructions.

45. PROBLEM. To find a point whose distances from the extremities of a given segment are specified.

SOLUTION. Let AB be the given segment and let it be required to find a

A

B

point C which shall be one inch from each extremity of AB.

Set the points of the compasses one inch apart. With A as a center draw an arc m, and with B as a center draw an arc n meeting the arc m in the point c. Then every point in the arc m is one inch from A and every point in the arc n is one inch from B (§ 12).

Hence C, which lies on both m and n, is one inch from A and also from B.