xterior ngle Inequality efinition: Given Î, the angles p, p, and p are called interior angles of the triangle. ny angle that forms a linear pair with an interior angle is called an exterior angle. In the the diagram below, point is such that **, and p is an exterior angle. orresponding to this exterior angle are the remote (or opposite ) interior angles p and p. Theorem (The xterior ngle Inequality): In any triangle, an exterior angle has greater measure than either of the remote interior angles. That is, given Î and point such that **,. μ( ) > μ( ) and μ( ) > μ( ) M ~ Given Î and point such that **: Find the midpoint M of and find point such that *M* and M = M. onstruct segment the same side of. y the X Theorem, and are on, and since both ** and *M*, and are on the same side of. Thus, point is interior to p, and we have. lso, since M = M, M = M, and the μ( ) + μ( ) = μ( ) vertical angles pm and pm are congruent, SS gives us ÎM ÎM. PF gives us µ(pm) = µ(pm). Thus µ(p) = µ(p) + µ(p) > µ(p) = µ(p). Now to show that µ(p) > µ(p), we utilize the work we just did. Find point F such that **F. Note that p and pf are vertical angles and are therefore congruent; µ(p) = µ(pf). ut then, the exact construction we have just done can be used to show µ(pf) > µ(p).
M N F G Find the midpoint N of, and G with *N*G and N = NG. We can show G is interior to pf, use SS to get ÎN ÎGN and PF to get µ(png) = µ(pn), and so µ(p) = µ(pf) = µ(pg) + µ(pgf) > µ(pg) = µ(p).
couple of corollaries not in out text: orollary 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). 2 3 1 orollary 2: triangle can have at most one right or obtuse angle. (n immediate consequence of orollary 1). orollary 3: ase angles of an isosceles triangle are acute. (n immediate consequence of orollary 2)
orollary (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. P l m Q n R S Now ppqr and pprq are right angles, as is pprs. ut 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 orollary 2 above, ΔPQR has two right angles.
Theorem (S ongruence): 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 Δ ΔXYZ with p px, p py, and XZ. Our strategy is to show p pz and apply S. So, WLOG, we assume for contradiction that µ(p) > µ(pz). onstruct ray P such that µ(pp) = µ(pz) and P between and. (How?) Now P is interior to p and so P meets at some point. (Why?) Now ª ªXYZ by S, and so µ(p) = µ(py), by PF. ut: µ(p) > µ(p) = µ(py), contradicting the exterior angle inequality. So p pz, and S completes the proof. -ish. Z P X Y 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.
Theorem (SSS ongruence): 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 Î and ÎXYZ under the correspondence : XYZ. Using P P the Protractor and Ruler Postulates, find a ray on the side of opposite from, and such that µ(pp) = µ(pzxy), and find a point on such that = XY. Note that since = XZ, = XY, and µ(p) = µ(pzxy), Î ÎXYZ by SS. X Y Z Since we created on the other side of from, the segment intersects the line at some point. There are five possibilities as to how relates to points and : =, =, **, **, or **. We address the case **, as shown in our diagram above. Note first that is a crossbar for both p and p. Thus is interior to both angles, and thus we know µ(p) + µ(p) = µ(p). Similarly, µ(p) + µ(p) = µ(p). Since = XY =, Î is isosceles. lso, since = YZ =, Î is isosceles. Thus, µ(p) = µ(p) and µ(p) = µ(p). So, µ(p) = µ(p) + µ(p) = µ(p) + µ(p) = µ(p). ut since Î ÎXYZ we know µ(p) = µ(pxyz). So µ(p) = µ(p) = µ(pxyz). This, together with knowing = XY and = YZ, allows us to use SS to conclude that Î and ÎXYZ are congruent. Note: The argument above depends on ** (where did we use that fact without explicitly mentioning it?). However, as noted above, it could be that ** or ** or = or =.
s 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. Y X Z