Transcript Congruence by S.A.S.

Side-Angle-Side Congruence
by basic rigid motions
A geometric realization of a proof in
H. Wu’s “Teaching Geometry According
to the Common Core Standards”
Given two triangles, ABC and A0B0C0.
Assume two pairs of equal
corresponding sides with
the angle between them
equal.
C
B0
side
C0
A
side
B
A0
We want to prove the triangles are congruent.
In other words, given ABC and A0B0C0, with
|AB| = |A0B0|,  A =  A0,
and |AC| = |A0C0|,
B0
C
side
C0
A
side
B
A0
we must give a composition of basic rigid
motions that maps ABC exactly onto A0B0C0.
We first move vertex A to A0 by a translation 𝑇
along the vector from A to A0
B0
C
A
C0
B
A0
𝑇 translates all points in the plane. Original positions
are shown with dashed lines and new positions in red.
Then we use a rotation ℛ to bring the horizontal
side of the red triangle (which is the translated
image of AB by 𝑇) to A0B0.
B0
C
A
C0
B
A0
ℛ maps the translated image of AB exactly onto
A0B0 because |AB| = |A0B0| and translations
B0
preserve length.
C
C0
A
B
A0
Now we have two of the red triangle’s vertices
coinciding with A0 and B0 of  A0B0C0.
B0
C
A
C0
B
A0
After a reflection of the red triangle across A0B0, the
third vertex will exactly coincide with C0.
Can we be sure this composition of basic rigid
motions (the reflection of the rotation of the
translation of the
B0
image of ABC)
C
A
C0
B
A0
takes C to C0 — and the red triangle with it?
Yes! The two marked angles at A0 are equal since
basic rigid motions preserve degrees of angles,
B0
and CAB =  C0A0B0
is given by hypothesis.
C
A
C0
B
A0
A reflection across A0B0 does take C to C0
— and the red triangle with it!
Since basic rigid motions preserve length
and since |AC| = |A0C0|,
B0
after a reflection
across A0B0,
C
A
C0
B
A0
by Lemma 8, the red triangle coincides with A0B0C0.
The triangles are congruent. Our proof is complete.
Given two triangles with two pairs of equal sides
and an included equal angle, a composition of
B0
basic rigid motions
(translation, rotation,
and
C
C0
reflection)
A
B
A0
maps the image of one triangle onto the other.
Therefore, the triangles are congruent.