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Integrated math 1

Lesson 2: Triangle congruence from transformations

Justify triangle congruence

Below are

△ A B C

and

△ D E F

. We assume that

A B = D E

B C = E F

, and

m ∠ B = m ∠ E

Here is a rough outline of a proof that

△ A B C ≅ △ D E F

We can map $△ A B C$ ‍ using a sequence of rigid transformations so that $A^{'} = D$ ‍ and $B^{'} = E$ ‍.
[Show drawing.]
If $C^{'}$ ‍ and $F$ ‍ are on the same side of $\overset{\leftrightarrow}{D E}$ ‍, then $C^{'} = F$ ‍.
[Show drawing.]
If $C^{'}$ ‍ and $F$ ‍ are on opposite sides of $\overset{\leftrightarrow}{D E}$ ‍, then we reflect $△ A^{'} B^{'} C^{'}$ ‍ across $\overset{\leftrightarrow}{D E}$ ‍ and then $C^{″} = F$ ‍, $A^{″} = D$ ‍ and $B^{″} = E$ ‍.
[Show drawing.]

What is the justification that $C^{'} = F$ ‍ in step 2?

(Choice A)
$C^{'}$ ‍ and $F$ ‍ are the same distance from $E$ ‍ along the same ray.
(Choice B)
Both $C^{'}$ ‍ and $F$ ‍ lie on intersection points of circles centered at $D$ ‍ and $E$ ‍ with radii $D F$ ‍ and $E F$ ‍, respectively. There are two such possible points, one on each side of $\overset{\leftrightarrow}{D E}$ ‍.
(Choice C)
$C^{'}$ ‍ and $F$ ‍ are at the intersection of the same pair of rays.

Stuck?