Geometry for Elementary School/The Side-Side-Side congruence theorem

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In this chapter, we will start the discussion of congruence and congruence theorems. We say the two triangles are congruent if they have the same shape. The triangles \triangle ABC and \triangle DEF are congruent if and only if all the following conditions hold:

  1. The side \overline {AB} equals \overline {DE} .
    Geom side congr 01.png


  2. The side \overline {BC} equals \overline {EF} .
    Geom side congr 02.png


  3. The side \overline {AC} equals \overline {DF} .
    Geom side congr 03.png


  4. The angle \angle ABC equals \angle DEF .
    Geom side congr 04.png


  5. The angle \angle BCA equals \angle EFD .
    Geom side congr 05.png


  6. The angle \angle CAB equals \angle FDE .
    Geom side congr 06.png


Note that the order of vertices is important. It is possible that \triangle ABC and \triangle ACB are not congruent even though both refer to the same triangle.

Congruence theorems give a set of the fewest conditions that are sufficient in order to show that two triangles are congruent.

The first congruence theorem we will discuss is the Side-Side-Side theorem.

Contents

[edit] The Side-Side-Side congruence theorem

Given two triangles \triangle ABC and \triangle DEF such that their sides are equal, hence:

  1. The side \overline {AB} equals \overline {DE} .
    Geom side conth 01.png


  2. The side \overline {BC} equals \overline {EF} .
    Geom side conth 02.png


  3. The side \overline {AC} equals \overline {DF} .
    Geom side conth 03.png


Then the triangles are congruent and their angles are equal too.
Geom side conth 04.png


[edit] Method of Proof

In order to prove the theorem we need a new postulate. The postulate is that one can move or flip any shape in the plane without changing it. In particular, one can move a triangle without changing its sides or angles. Note that this postulate is true in plane geometry but not in general. If one considers geometry over a ball, the postulate is no longer true.


Given the postulate, we will show how can we move one triangle to the other triangle location and show that they coincide. Due to that, the triangles are equal.

[edit] The construction

  1. Copy The line Segment side \overline {AB} to the point D.
  2. Draw the circle \circ D,\overline{AB} .
  3. The circle \circ D,\overline{AB} and the segment \overline{DE} intersect at the point E hence we have a copy of \overline {AB} such that it coincides with \overline{DE} .
  4. Construct a triangle with \overline{DE} as its base, \overline{BC} , \overline{AC} as the sides and the vertex at the side of the vertex F. Call this triangle triangles \triangle DEG

[edit] The claim

The triangles \triangle DEF and \triangle ABC congruent.


[edit] The proof

  1. The points A and D coincide.
  2. The points B and E coincide.
  3. The vertex F is an intersection point of \circ D,\overline{DF} and \circ E,\overline{EF} .
  4. The vertex G is an intersection point of \circ D,\overline{AC} and \circ E,\overline{BC} .
  5. It is given that \overline{DF} equals \overline{AC} .
  6. It is given that \overline{EF} equals \overline{BC} .
  7. Therefore, \circ D,\overline{DF} equals \circ D,\overline{AC} and \circ E,\overline{EF} equals \circ E,\overline{BC} .
  8. However, circles of different centers have at most one intersection point in one side of the segment that joins their centers.
  9. Hence, the points G and F coincide.
  10. There is only a single straight line between two points, therefore \overline {EG} coincides with \overline {EF} and \overline {GD} coincides with \overline {DF} .
  11. Therefore, the \triangle DEG coincides with \triangle DEF and the two are congruent.
  12. Due to the postulate \triangle DEG and \triangle ABC are equal and therefore congruent.
  13. Hence, \triangle DEF and \triangle ABC are congruent.
  14. Hence, \angle ABC equals \angle DEF , \angle BCA equals \angle EFD and \angle CAB equals \angle FDE .

[edit] Note

The Side-Side-Side congruence theorem appears as Book I, prop 8 at the Elements. The proof here is in the spirit of the original proof. In the original proof Euclid claims that the vertices F and G must coincide but doesn’t show why. We used the assumption that “circles of different centers have at most one intersection point in one side of a segment that joins their centers”. This assumption is true in plane geometry but doesn’t follows from Euclid’s original postulates. Since Euclid himself had to use such an assumption, we preferred to give a more detailed proof, though the extra assumption.

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