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Congruence Theorems

According to our development, congruence has been a relationship between segments or a relationship between angles. In Geometry we are accustomed to seeing congruence as a relationship between triangles. We can make this so by a definition.

Definition: Two triangles are congruent  if there exists a one-to-one correspondence between their vertices sot that the corresponding sides and corresponding angles are congruent.

Proposition 7.1:  If in tex2html_wrap_inline11270 we have tex2html_wrap_inline13516, then tex2html_wrap_inline12628.

 figure1576
Figure 8.1: Isosceles triangles 

Proof: This simple proof is due to Pappus. Consider the correspondence of vertices tex2html_wrap_inline13526, tex2html_wrap_inline13528, and tex2html_wrap_inline13530. Under this correspondence, two sides and the included angle of tex2html_wrap_inline11270 are congruent respectively to the corresponding sides and included angle of tex2html_wrap_inline13534. Hence, by SAS the triangles are congruent. Therefore, the corresponding angles are congruent and tex2html_wrap_inline12628.

Proposition 7.2:[Segment Subtraction]   If tex2html_wrap_inline12498, tex2html_wrap_inline13540, tex2html_wrap_inline13542, and tex2html_wrap_inline13544, then tex2html_wrap_inline13546.

 figure1596
Figure 8.2: Segment subtraction 

Proof: From what we are given, let us assume that tex2html_wrap_inline13560. By Axiom C-1, there exists a unique point G on ray tex2html_wrap_inline13564 so that tex2html_wrap_inline13566. By our hypothesis, we have that tex2html_wrap_inline13568. Since tex2html_wrap_inline13542 and tex2html_wrap_inline13566, by Axiom C-3 tex2html_wrap_inline13574. But then by Axiom C-2 we have that tex2html_wrap_inline13576. It follows from Axiom C-1 that F=G. We have reached a contradiction. Thus, it must be that tex2html_wrap_inline13546, and we are done.

Proposition 7.3:  Given tex2html_wrap_inline13544, then for any point B between A and C, there exists a unique point E, tex2html_wrap_inline13540, such that tex2html_wrap_inline13542.

Proof: The proof of this proposition is left to the reader.

We can use this result to help us establish a partial ordering on the line segments in the plane.

Definition: AB < CD (or CD > AB) means that there exists a point E between C and D such that tex2html_wrap_inline13606. 

Proposition 7.4:[Segment Ordering] 

Proof: We will prove the first item only. The proofs of the remaining three items will be left to you as homework.

We are given segments AB and CD. If tex2html_wrap_inline12592, then we are done. So, let us assume that tex2html_wrap_inline13632. By Axiom C-1 there is a unique point tex2html_wrap_inline13634 so that tex2html_wrap_inline13606. tex2html_wrap_inline13638, else tex2html_wrap_inline13640 which is impossible under our assumption. Thus, we must have that tex2html_wrap_inline13642 or tex2html_wrap_inline13644 from Axiom B-2. If tex2html_wrap_inline13642, then by definition AB < CD, and we are done.

Suppose then that tex2html_wrap_inline13644. By Proposition 8.3 there exists a unique point tex2html_wrap_inline13652 so that tex2html_wrap_inline13654. In this case by the definition AB > CD, and we are done.

Proposition 7.5:  Supplements of congruent angles are congruent.

Proposition 7.6: 

Proposition 7.7:  For every line tex2html_wrap_inline11154 and every point P there exists a line through P perpendicular to tex2html_wrap_inline11154.

 figure1661
Figure: Theorem 8.7 

Proof: Either tex2html_wrap_inline13680 or tex2html_wrap_inline13682. First, let us assume that tex2html_wrap_inline13682. Let tex2html_wrap_inline13686. Such points exist by Axiom I-2. The ray tex2html_wrap_inline13688 lies on one side of tex2html_wrap_inline11154. By Axiom C-4 there exists a ray tex2html_wrap_inline13692 on the opposite side of tex2html_wrap_inline11154 from P so that
displaymath13507
By Axiom C-1 there is a point tex2html_wrap_inline13698 so that tex2html_wrap_inline13700. Since P and P' are on opposite sides of tex2html_wrap_inline11154, the segment PP' intersects the line tex2html_wrap_inline11154. Let tex2html_wrap_inline13712.

If Q=A, then tex2html_wrap_inline13716. Thus, tex2html_wrap_inline13718.

If tex2html_wrap_inline13720, then tex2html_wrap_inline13722 by SAS. Thus, from the definition of congruent triangles, tex2html_wrap_inline13724, and tex2html_wrap_inline13718.

Now, if tex2html_wrap_inline13680, there is an tex2html_wrap_inline13730. From this point X apply the previous technique to construct a perpendicular line to tex2html_wrap_inline11154 through X. By Axiom C-4 we can copy this angle on one side of tex2html_wrap_inline11154 at P. From the second part of Proposition 8.6 the other side of this angle is part of a line through P perpendicular to tex2html_wrap_inline11154.

Proposition 7.8:[ASA]  Given tex2html_wrap_inline11270 and tex2html_wrap_inline13748 with tex2html_wrap_inline13750, tex2html_wrap_inline13752, and tex2html_wrap_inline13544. Then tex2html_wrap_inline13756.

Proposition 7.9:  If in tex2html_wrap_inline11270 we have that tex2html_wrap_inline12628, then tex2html_wrap_inline13516 and tex2html_wrap_inline11270 is isosceles.

Proposition 7.10:[Angle Addition]   Given tex2html_wrap_inline13766 between tex2html_wrap_inline13314 and tex2html_wrap_inline13316, tex2html_wrap_inline13772 between tex2html_wrap_inline13774 and tex2html_wrap_inline13776, tex2html_wrap_inline13778 and tex2html_wrap_inline13780. Then tex2html_wrap_inline13782.

Proposition 7.11:[Angle Subtraction]   Given tex2html_wrap_inline13766 between tex2html_wrap_inline13314 and tex2html_wrap_inline13316, tex2html_wrap_inline13772 between tex2html_wrap_inline13774 and tex2html_wrap_inline13564, tex2html_wrap_inline13778, and tex2html_wrap_inline13780, then tex2html_wrap_inline13782.

 figure1718
Figure 8.4: Angle subtraction 

Proof: Since tex2html_wrap_inline13766 lies between tex2html_wrap_inline13314 and tex2html_wrap_inline13316, we may apply the Crossbar Theorem to find that tex2html_wrap_inline13766 intersects AC. Without loss of generality, we may assume that this point of intersection is, in fact, the point G. Then, we have that tex2html_wrap_inline13832. Assume that the points D, F, and H are chosen so that tex2html_wrap_inline13840, tex2html_wrap_inline13842, and tex2html_wrap_inline13546. This is nothing but a relabeling of the points.

Let us assume then that tex2html_wrap_inline13846. Then, there exists a unique ray, tex2html_wrap_inline13848, on the same side of tex2html_wrap_inline13850 as tex2html_wrap_inline13774 so that
displaymath13508
By our assumption, tex2html_wrap_inline13854, so by Proposition 8.10, tex2html_wrap_inline13856. Since tex2html_wrap_inline13782, the uniqueness of tex2html_wrap_inline13848 implies that tex2html_wrap_inline13862, a contradiction. Thus, tex2html_wrap_inline13864.

As with line segments, there is a natural method for defining an ordering on angles.

Definition: tex2html_wrap_inline13866 means that there exists a ray tex2html_wrap_inline13868 between tex2html_wrap_inline13774 and tex2html_wrap_inline13564 so that tex2html_wrap_inline13874. 

This gives us the following results, which are completely analogous to those for segments.

Proposition 7.12:[Ordering of Angles] 

Proposition 7.13:[SSS]   Given triangles tex2html_wrap_inline11270 and tex2html_wrap_inline13748. If tex2html_wrap_inline13542, tex2html_wrap_inline13546, and tex2html_wrap_inline13544, then tex2html_wrap_inline13756.

Proposition 7.14:[Euclid's Fourth Postulate]  All right angles are congruent to each other.


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