Download Theorem: AAS Congruence. If under some correspondence, two

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 ªABC : ªXYZ, with pA –pX, pB–pY, and
Our strategy is to show pC–pZ and apply ASA. So,
WLOG, we assume for contradiction that mpC > mpZ.
Construct ray
such that mpACP = mpZ and
. (How?) Now P is interior to pACB and so
meets
at some point D. (Why?) Now ªADC – ªXYZ
by ASA, and so mpADC =mpY, by CPCF. But: mpADC >
mpB = mpY, contradicting the exterior angle inequality. So
pC–pZ, and ASA completes the proof.
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, pCAB –pZXY, AC = XZ, and CB = ZY, but clearly the
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.
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 ªABC and
ªXYZ with pCAB –pZXY, AC = XZ, and CB = ZY, but ABC
† ªXYZ; we show pABC and pXYZ are supplementary.
If pACB /pXZY, then the triangles are congruent by ASA. So,
WLOG, mpACB > mpXZY. Find ray
with
and
mpACP = mpXZY.
intersects
at some point D with A-DB. By ASA, ªADC / ªXYZ. By CPCF, pADC / pXYZ and CD =
ZY. Since pADC and pCDB are a linear pair, they are
supplementary, so mpADC + mpCDB = 180, so mpXYZ + mpCDB
= 180. Finally, since CB = ZY = CD, ªCDB is isosceles so mpABC
= mpCDB, so we have mpXYZ + mpABC =180.
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.
Theorem: SsA Congruence: Given ªABC and ªXYZ, suppose
pCAB –pZXY, AC = XZ, CB = ZY, and CB > CA. Then the two
triangles are congruent.
Proof: 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.