Answer:
ΔA'B'C' is congruent to ΔABC
Step-by-step explanation:
See attached for the construction of ΔA'B'C'. (The vertices are labeled ABC.)
We computed angle B to be ...
∠B = 180° -∠A -∠C
∠B = 180° -103° -57° = 20°
We constructed segment BC of length 6.5. Then we constructed angles of 20° and 57° from B and C, respectively. The location where the rays from those angles cross is point A', and the angle there is 103°, as required.
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ΔA'B'C' is congruent to ΔABC by the ASA congruence postulate.