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Transcript
A Digression into SSA or, as the Textbook Prefers, ASS 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. For example, referring to the diagram below, pCAB pZXY, AC = XZ, and CB = ZY, but clearly ªABC is not congruent to ªXYZ. 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 pB and pY, and 2) make some restrictions that will guarantee congruence in some cases. Theorem (SSA Theorem - Not in the Text): 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, then the remaining pair of angles not included by the congruent sides are supplementary. ~ The proof reduces to SAS if the angle is included between the two sides, so assume otherwise. Given ªABC and ªXYZ with pCAB pZXY, AC = XZ, and CB = ZY, but ªABC á ªXYZ; we show pABC and pXYZ are supplementary. If pACB pXZY, then the triangles are congruent by ASA. So, WLOG, µ(pACB) > µ(pXZY). Find ray and with µ(pACP) = µ(pXZY). between intersects and at some point D with A*D*B. By ASA, ªADC / ªXYZ. By CPCF, pADC / pXYZ and CD = ZY. Since pADC and pCDB are a linear pair, they are supplementary, so µ(pADC) + µ(pCDB) = 180, so µ(pXYZ) + µ(pCDB) = 180. Finally, since CB = ZY = CD, ªCDB is isosceles so µ(pABC) = µ(pCDB), so we have µ(pXYZ) + µ(pABC) =180. Some Easy Corollaries: Corollary A: 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. ~ 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. Corollary B (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. ~ 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. Corollary C (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. ~ Follows immediately from AAS Theorem. Corollary D (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. ~ Follows immediately from AAS Theorem. Cororllary E (SsA Congruence Theorem – Not in the Text): Given ªABC and ªXYZ, suppose pCAB pZXY, AC = XZ, CB = ZY, and CB > CA. Then the two triangles are congruent. ~ If pCAB pZXY, AC = XZ, CB = ZY, then by the SSA Theorem, if the triangles are not congruent, pB and pY 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. Intuitively, this is the case where is longer than and so cannot “pivot around” the point C to come to rest on two different places on the side opposite C.