BOOK I

PROPOSITION I. THEOREM

30. If two triangles have two sides and the included angle of one equal respectively to two sides and the included angle of the other, the triangles are equal in all respects.

B

E

A

Let the AABC and DEF have AB = DE, BC= LB=LE.

To Prove the ▲ ABC and DEF equal in all respects.

EF, and

Proof. Place the ▲ ABC upon the ▲ DEF so that B shall coincide with its equal E, BA falling upon ED, and BC upon

EF.

=

Since, by hypothesis, BA ED, the vertex 4 will fall upon the vertex D.

Since, by hypothesis, BC EF, the vertex C will fall upon the vertex F.

coin

Since, by Axiom 13, only one straight line can be drawn joining two points, 4C will coincide with DF. .. the cide throughout and are equal in all respects.

Q.E.D.

31. SCHOLIUM. By showing that the A coincide, we have not only proved that they are equal in area, but also that LA=LD, LC = ≤ F, and AC = DF.

It should be noticed that the sides AC and DF, which have been proved equal, lie opposite respectively to the equal angles B and E.

Also, that the equal angles 4 and D lie opposite respectively to the equal sides BC and EF, and that the equal angles C and F lie opposite respectively to the equal sides AB and DE.

PRINCIPLE. In triangles that have been proved equal in all respects, equal sides lie opposite equal angles, and equal angles lie opposite equal sides.

Suggestion. Show by § 30 that the AABD and ADC are equal in all respects. Then, by the principle of § 31, BD = DC.

34. EXERCISE. ABC is a triangle having AB BC. BE is laid off equal to BD.

A

PROPOSITION II. THEOREM

35. 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 equal in all respects.

B

A

E

A

D

Let the AABC and DEF have ▲ A = LD, LC = LF, and

AC DF.

To Prove the AABC and DEF equal in all respects.

Proof. Place the ▲ ABC upon the ▲ DEF, so that ▲▲ shall coincide with its equal D, AB falling upon DE, and AC

falling upon DF.

Since, by hypothesis, ACDF, the vertex C will fall upon

vertex F.

Since, by hypothesis, C=F, the side CB will fall upon FE, and the vertex B will be on FE or its prolongation.

Since AB falls upon DE, the vertex B will be upon DE or its prolongation.

The vertex B, being at the same time on DE and FE, must be at their point of intersection; and since two straight lines have only one point of intersection (Axiom 13), the vertex B must fall at E.

.. the

ABC and DEF coincide throughout, and are equal

in all respects.

Q.E.D.

SANDERS' GEOM. -2

36. EXERCISE. Prove Prop. II., using this pair of triangles.

37. EXERCISE. In the ▲ ABC, BD bisects ZABC and is perpendicular to AC.

Prove that BD bisects AC and that AB

= BC.

38. EXERCISE. ABC is a ▲ having ▲ BAC A

= BCA. AD bisects BAC and CE bisects LBCA.

Suggestion. Prove & ADC and AEC equal

in all respects by § 35. Then by the Principle

of § 31, AD = EC.

39. The next proposition is an example of what is called the indirect proof.

The reasoning is based on the following Principle: If the direct consequences of a certain supposition are false, the supposition itself is false.

To prove a theorem by this plan, the following steps are necessary:

1. The theorem is supposed to be untrue.

2. The consequences of this supposition are shown to be false.

3. Then, by the above Principle, the supposition (that the theorem is untrue) is false.

4. The theorem is therefore true.