Proving SAS with Reflections — showing two triangles are congruent (using points V, M, R, N, C, F)

We prove the Side–Angle–Side (SAS) congruence statement by using reflections. The pictures in the worked example show two triangles that share the vertex V and have two pairs of corresponding sides equal and the included angle equal. The idea is: place one triangle on top of the other by reflecting across appropriate lines. Reflections preserve distances and angles, so when you can place the two corresponding sides and included angle to coincide, the whole triangles must coincide.

Step-by-step idea (generic):

• Suppose triangles ΔVAB and ΔVA'B' satisfy VA = VA' and VB = VB' and ∠AVB = ∠A'VB'. (In the diagram the labels correspond to V, M, R, N, C, F — the role of A, B, A', B' is played by those points.)
• Because VA = VA', point A' must lie on the circle centered at V with radius VA. Similarly, because VB = VB', point B' must lie on the circle centered at V with radius VB. So each corresponding vertex of the second triangle is constrained to a circle about V.
• For a fixed position of the rays that determine the included angle at V, each vertex constrained to its circle has at most two possible positions (symmetric about the bisector). Thus there are at most two symmetric placements of the second triangle relative to the first.
• If the second triangle is not already exactly on top of the first, reflect across the perpendicular bisector of the segment joining the two corresponding vertices (that perpendicular bisector passes through V because those two vertices are equidistant from V). That reflection swaps those two vertices while keeping V fixed and preserves the angle at V, so the reflected triangle now matches the first triangle exactly.
• Because reflections preserve all lengths and angle measures, once the two sides and the included angle of one triangle match the corresponding two sides and included angle of the other triangle, every side and every angle of the first triangle matches the corresponding side and angle of the second. Hence the triangles are congruent (SAS).

Now the specific missing pieces and short answers (with reasoning):

1. Sentence: "Point ___ can be at only two places on the circle because ___ must be congruent to ___."
   Answer: "Point F can be at only two places on the circle because VF must be congruent to VM."
   Reason: If VM = VF then F lies on the circle centered at V with radius VM. For a fixed ray (or fixed included angle at V) there are only two symmetric intersection points of that ray and the circle, so F has only two possible positions.
2. Sentence: "Point m and point ___ are equidistant from point v, so v lies on the perpendicular bisector of ___. This is why a reflection across ___ maps point ___ onto point m."
   Answer: "Point m and point F are equidistant from point v, so v lies on the perpendicular bisector of MF. This is why a reflection across the perpendicular bisector of MF maps point F onto point m."
   Reason: If VM = VF then V is equidistant from M and F, so V lies on the perpendicular bisector of segment MF; reflecting across that perpendicular bisector swaps M and F while fixing V.
3. Radio/short conclusion: "Because a series of reflections preserves distance and angle measures, ___ congruent to ___."
   Answer: "ΔVFR is congruent to ΔVMR."
   Reason: After the appropriate reflection(s) we have made the corresponding sides and included angle coincide; reflections preserve segment lengths and angle measures, so the whole triangle is carried to the other triangle — they are congruent. (Any equivalent matching pair of triangles from the diagram can be stated here; the key is that the triangle before reflection is congruent to the triangle after reflection.)
4. Second mirror sentences (same logic applied to the other vertex):
5. Sentence: "Point ___ must be on the circle centered at v with radius v times r, because ___."
   Answer: "Point R must be on the circle centered at V with radius VR, because RR is congruent to VR."
   (Interpretation: the intended meaning is that the corresponding vertex must lie on the circle about V whose radius equals the distance from V to the corresponding vertex in the first triangle — so that the two corresponding side-lengths are equal.)
6. Sentence: "If point ___ is not at r, then a reflection across ___ will map point ___ onto point r."
   Answer: "If point R' is not at R, then a reflection across the perpendicular bisector of RR' will map point R' onto point R."
   Reason: As before, reflecting across the perpendicular bisector of the segment joining the two candidate positions for the corresponding vertex swaps those two positions and fixes V, so it places that vertex where it must be.

Final conclusion: By reflecting one triangle (or composing at most two reflections) we can place its two given congruent sides and included angle exactly on the corresponding sides and included angle of the other triangle. Reflections preserve distances and angle measures, so the entire triangles coincide after the reflection(s). Therefore triangles with two pairs of corresponding sides and the included angle congruent are congruent — this is the Side–Angle–Side (SAS) Congruence Theorem.

If you want, I can rewrite the blanks using the exact label choices shown in your diagram (explicitly matching V, M, R, N, C, F) — tell me which two triangles you are comparing and I will fill the blanks with those exact labels.