Download Similarity Definition: Two triangles and are said to be similar

Similarity
Definition: Two triangles and
are said to be similar, denoted
, iff under the usual correspondence of their vertices
(A :X, B : Y, C :Z), corresponding angles are congruent.
Theorem (Wallis’s Postulate): If
is a triangle and
segment, then there exists a point F such that
~ Use the protractor postulate to copy construct a ray
µ(pEDG) = µ(pCAB) and another ray
is a
.
such that
such that H and G are on
the same side of
and µ(pDEH) = µ(pABC). By Euclid’s
Postulate V, since these two angles have measures that add to less than
180, the rays
and
sums for
and for
µ(pC) = µ(pF). Then
must meet at a point F. Since the angle
are both 180, it follows that
. €
Note: Wallis’s Postulate is actually equivalent to the Euclidean
Parallel Postulate – the above theorem proves one direction, and the
other is proved on page 123 of your text.
Note also that this definition of similarity is somewhat unusual;
typically additional requirements involving the ratios of sides are also
included. Here, we prove such requirements follow from our
definition. This has the advantage of being able to talk about
similarity in non-Euclidean geometry (although we will see it reduces
to congruence in hyperbolic geometry). It also makes the proof of the
following theorem immediate:
Theorem (AA Similarity Criterion): If two angles of one triangle are
congruent to two angles of a second triangle, the triangles are similar.
Theorem (Similar Triangles Theorem): If
then
.
~ If AB = XY, the triangles are congruent by ASA, so AB = XY and
AC = XZ and the result if trivial. WLOG, assume AB > XY. Locate
D such that A*D*B and AD = XY. Construct a line parallel to
that intersects
at E (Note the Pasch gives us that E exists). The
Parallel Projection Theorem establishes that the measures of the sides
are in the same ratio, so
. Moreover, corresponding
angles give us that ªADE –
XZ. Thus
ªXYZ , so we have AD = XY and AE =
, and algebra gives
.€
Theorem: If
, then
.
~ As above, if AB = XY, the triangles are congruent by ASA, so AB =
XY and AC = XZ and BC = YZ and the result if trivial. WLOG,
assume AB > XY. Locate D such that
A*D*B and AD = XY. Construct a line
parallel to
that intersects
at E
(Note the Pasch gives us that E exists). The
Parallel Projection Theorem establishes that
the measures of the sides are in the same
ratio, so
. To get the second
equality, construct segment
parallel to
. Then, by the Side-Splitting Theorem
again,
. We do some algebra:
(Adding 1 to both sides)
(Getting a common denominator)
(Using betweenness, AC =AE+EC and
BC = BF + FC)
(Since BDEF is a parallelogram,
DE = FB)
Taking reciprocals and re-writing.
Finally, corresponding angles give us that ªADE –
have AD = XY, AE = XZ, and DE = YZ. Thus
which completes the proof. €
ªXYZ , so we
,
is called the common
Note: This number,
ratio of the sides of the similar triangles.
Theorem (SAS Similarity Criterion): If in ªABC and ªXYZ we
have
~ If
and pA–pX, then ªABC -ªXYZ.
, then AB = XY and AC = XZ and SAS gives us
that the two triangles are congruent, and thus similar. So, assume the
common ratio is not 1, and so either AB or XY is larger. WLOG, let
AB > XY. Then AC > XZ as well. On
find point D such that AD-B and AD = XY, and on AC find point E such that A*E*C and AE
= XZ. Note first that ªADE – ª XYZ by SAS. Next, note that
and we have
by the Lemma of the Parallel
Projection Theorem. By transversal properties and CPCF, pB – pD –
pY, and pE – pC – pZ, so the corresponding angles are all the same.
€
Theorem (SSS Similarity Criterion): If in ªABC and ªXYZ we
, then pA – pX, pB – pY, pC – pZ, and so
have
ªABC -ªXYZ.
~ If the common ratio is one, the triangles are congruent, so WLOG
assume AB, BC, and AC are the longer sides. On
and
,
respectively, locate points D and E and such that B*D*A, B*E*C, BD
= XY and BE = YZ. Then
, so by the
lemma of the last section,
we then have
. By the side splitting theorem
. But
, so we have
, or DE = XY. We can then conclude that ªXYZ
–ªDBE by SSS. We have immediately that pABC –pDBE –pXYZ.
And, by transversal theorems, pBDE –pBAC and pBED – pBCA.
Thus by CPCF, pYXZ –pBAC and pYZX – pBCA. €