Download Exterior Angle Inequality, AAS

Exterior Angle Inequality
Definition: Given ÎABC, the angles pACB, pCBA, and pBAC are
called interior angles of the triangle. Any angle that forms a linear pair
with an interior angle is called an exterior angle. In the the diagram
below, point D is such that A*C*D, and pBCD is an exterior angle.
Corresponding to this exterior angle are the remote (or opposite )
interior angles pABC and pCAB.
Theorem (The Exterior Angle Inequality): In any triangle, an
exterior angle has greater measure than either of the remote interior
angles. That is, given ÎABC and point D such that A*C*D,
.
~ Given ÎABC and point D such that A*C*D: Find the midpoint M
of
and find point E such that A*M*E and AM = ME. Construct
segment
. By the X Theorem, E and B are on the same side of
, and since both A*C*D and A*M*E, E and D are on the same
. Thus, point E is interior to pBCD, and we have
. Also, since
BM = CM, AM = EM, and the vertical angles pBMA and pCME are
congruent, SAS gives us ÎBMA –ÎCME. CPCF gives us
µ(pMBA) = µ(pMCE). Thus µ(pBCD) = µ(pBCE) + µ(pECD) >
µ(pBCE) = µ(pABC).
side of
Now to show that µ(pBCD) > µ(pCAB), we utilize the work we just
did. Find point F such that B*C*F. Note that pBCD and pACF are
vertical angles and are therefore congruent; µ(pBCD) = µ(pACF). But
then, the exact construction we have just done can be used to show
µ(pACF) > µ(pCAB).
Find the midpoint N of
, and G with B*N*G and BN = NG. We
can show G is interior to pACF, use SAS to get ÎBAN –ÎGCN and
CPCF to get µ(pNCG) = µ(pNAB), and so µ(pBCD) = µ(pACF) =
µ(pACG) + µ(pGCF) > µ(pACG) = µ(pCAB). €
A couple of corollaries not in out text:
Corollary 1: The sum of the measures of any two angles of a triangle
is less than 180.
~ Given two angles we will call angles p1 and p2. Form the exterior
angle supplementary to p2; then 180 = µ(p2) + µ(p3) > µ(p2) +
µ(p1). €
Corollary 2: A triangle can have at most one right or obtuse angle.
(An immediate consequence of Corollary 1).
Corollary 3: Base angles of an isosceles triangle are acute. (An
immediate consequence of Corollary 2)
Corollary (Uniqueness of Perpendiculars): For every line l and for
every point P external to l, there exists exactly one line m such that P
is on m and l z m.
~ We proved in the last section that such a perpendicular line exists; it
remains to show that it is unique. We call that line m and the point
where l and m intersect point Q.
Now suppose for contradiction that there is another line n with P on n
and n z l. Let R be the point of intersection of l and n.
Now pPQR and pPRQ are right angles, as is pPRS. But pPRS is an
exterior angle for the triangle, and must be strictly larger than pPQR,
which it is not. This contradiction establishes the theorem. €
Note: For a different contradiction, we could have noted that, contrary
to Corollary 2 above,
has two right angles.
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.
~ (Outline of proof; you fill in the details as part of homework
problem 6.6.) Given the correspondence
with pA
–pX, pB–pY, and
. Our strategy is to show pC–pZ and
apply ASA. So, WLOG, we assume for contradiction that µ(pC) >
µ(pZ). Construct ray
such that µ(pACP) = µ(pZ) and
between
and
. (How?) Now P is interior to pACB and
so
meets
at some point D. (Why?) Now ªADC – ªXYZ by
ASA, and so µ(pADC) = µ(pY), by CPCF. But: µ(pADC) > µ(pB) =
µ(pY), contradicting the exterior angle inequality. So pC–pZ, and
ASA completes the proof. €-ish.
The book chooses this point to discuss the Hypotenuse-Leg (HL)
Theorem, but I prefer to do it later.
The last theorem of this section is the SSS theorem, which the book
proves by leaving it to you as an exercise. It can be proved as a
consequence of the triangle inequalities in the next section. Here, I’ll
prove it using the method suggested in the textbook, which is made
simpler by a simple lemma we talked about in class.
Lemma: The set of all points equidistant from each of two points A
and B is the perpendicular bisector of
.
~ This proof is a straightforward application of isosceles triangles and
congruence theorems, and makes a good exercise. Be sure to prove
both that if a point P is equidistant from A and B it is on the
perpendicular, and that if P is a point on the perpendicular bisector it is
equidistant from A and B. €-ish.
Theorem (SSS Congruence): If under some correspondence, three
sides of one triangle are congruent to the corresponding three sides of
another triangle, then the two triangles are congruent under that
correspondence.
~ Let the two triangles be ÎABC and ÎXYZ under the
correspondence ABC : XYZ. Using the Protractor and Ruler
Postulates, find a ray
on the side of
opposite from B, and
such that µ(pCAP) = µ(pZXY), and find a point D on
such that
AD = XY. Note that since AC = XZ, AD = XY, and µ(pCAD) =
µ(pZXY), ÎADC – ÎXYZ by SAS.
Now, since AD = XY = AB, A is equidistant from B and D. Also,
since BC = YZ = DC, C is also equidistant from B and D. Thus both
A and C are on the perpendicular bisector of
perpendicular bisector of
, so
is the
. In ÎABD, since AE = AE, µ(pAEB)
= 90 = µ(pAED), and BE = DE, SAS gives us ÎABE – ÎADE and
CPCF gives us pBAE – pDAE. This congruence, along with AB =
AD and AC = AC, allows us to use SAS to get ÎABC – ÎADC, and
since ÎADC – ÎXYZ we have ÎABC – ÎXYZ. €
Note: The argument above depends on A*E*C (where did we use that
fact without mentioning it?). However, it could be that A*C*E or
E*A*C. It also could be that A = E or C = E. As an exercise,
complete the proof for these cases, one of which is illustrated below.
This proof of SSS uses what are called “kites” and “darts” which are
the shapes of the objects constructed by putting the two triangles
together. The proof using triangle inequalities is actually shorter.