Proof: Given ABC XYZ, with A X, B Y, and Our strategy is to show C Z and apply ASA. So, WLOG, we assume for contradiction that m C > m Z.

Transcription

1 Theorem: AAS Congruence. If under some correspondence, two angles and a side opposite one of the angles of one triangle are congruent, respectively, to the corresponding two angles and side of a second triangle, then the triangles are congruent. Proof: Given ABC XYZ, with A X, B Y, and Our strategy is to show C Z and apply ASA. So, WLOG, we assume for contradiction that m C > m Z. Construct ray such that m ACP = m Z and. (How?) Now P is interior to ACB and so meets at some point D. (Why?) Now ADC XYZ by ASA, and so m ADC =m Y, by CPCF. But: m ADC > m B = m Y, contradicting the exterior angle inequality. So C Z, and ASA completes the proof.

2 Recall the ambiguous case from trigonometry: Here, you were asked to solve a triangle given two sides and an angle opposite one side. Before applying the law of sines, you needed to check to see how many triangles you had none, one, or two. Thus the information you were given did not uniquely determine a single triangle. Here, CAB ZXY, AC = XZ, and CB = ZY, but clearly the ABC is not congruent to XYZ. Thus the congruence of two sides and a non-included angle of one triangle to the corresponding two sides and non-included angle of another triangle is not enough to guarantee congruence of the triangles. However, we can 1) say some things about the relation between B and Y, and 2) make some restrictions that will guarantee congruence in some cases.

3 SSA Theorem: If, under some correspondence between their vertices, two triangles have two pairs of corresponding sides and a pair of corresponding angles congruent, and if the triangles are not congruent under this correspondence, the the remaining pair of angles not included by the congruent sides are supplementary. Proof: The proof reduces to SAS if the angle is included between the two sides, so assume otherwise. Given ABC and XYZ with CAB ZXY, AC = XZ, and CB = ZY, but ABC XYZ; we show ABC and XYZ are supplementary. If ACB XZY, then the triangles are congruent by ASA. So, WLOG, m ACB > m XZY. Find ray with and m ACP = m XZY. intersects at some point D with A-D- B. By ASA, ADC XYZ. By CPCF, ADC XYZ and CD = ZY. Since ADC and CDB are a linear pair, they are supplementary, so m ADC + m CDB = 180, so m XYZ + m CDB = 180. Finally, since CB = ZY = CD, CDB is isosceles so m ABC = m CDB, so we have m XYZ + m ABC =180.

4 Some Easy Corollaries: Theorem: If, under some correspondence between their vertices, two acute triangles have two sides and an angle opposite one of them congruent, respectively, to the corresponding two sides and angle of the other, the triangles are congruent. Proof: The hypothesis of this theorem satisfies that of the SSA theorem, so if the triangles are not congruent, the two remaining angles are supplementary. But this cannot be if all angles in both triangles are acute. So the triangles must be congruent. HL Theorem: If the hypotenuse and leg of one right triangle are congruent, respectively, to the hypotenuse and leg of a second right triangle, the two triangles are congruent. Proof: Since the hypothesis of this theorem satisfies that of the SSA theorem, if the two triangles are not congruent, the remaining (nonright) angles must be supplementary. However, in a right triangle, these angles must be acute, and so cannot be supplementary. Thus the triangles are congruent. HA Theorem: If the hypotenuse and one acute angle of one right triangle are congruent, respectively, to the hypotenuse and acute angle of a second right triangle, the two triangles are congruent. LA Theorem: If a leg and one acute angles of one right triangle are congruent, respectively, to a leg and acute angle of a second right triangle, the two triangles are congruent. Proof: Since right angles are congruent, these both follow directly from AAS.

5 Theorem: SsA Congruence: Given ABC and XYZ, suppose CAB ZXY, AC = XZ, CB = ZY, and CB > CA. Then the two triangles are congruent. Proof: If CAB ZXY, AC = XZ, CB = ZY, then by the SSA Theorem, if the triangles are not congruent, B and Y are supplementary. Thus, either they are both right angles or one is obtuse. In either case, they must be opposite the longest side of the triangle. But ZY = CB > CA = XZ, a contradiction. Thus the triangles must be congruent.