Answer:
I guess D is the answer for the question
so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.
in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B7%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B%5Cfrac%7B19%7D%7B2%7D%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D-7%20%5Cright%29%5E2%2B%5Cleft%28%20%5Cfrac%7B7%7D%7B2%7D-4%20%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B5%7D%7B2%7D%5Cright%29%5E2%2B%5Cleft%28%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Cboxed%7BAM%5Capprox%202.549509756796392%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Graph C shows the line of slope
If triangle ABC is congruent to triangle DEF, then EF = BC = 27
This is because BC and EF are the last two letters of ABC and DEF respectively. They match up and correspond, being congruent by CPCTC
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Similarly, if triangle ABC is congruent to triangle DEF, then angle D = angle A = 49 degrees
The letter D and the letter A are the first letters of DEF and ABC respectively. So they match up and are congruent by CPCTC
CPCTC = Corresponding Parts of Congruent Triangles are Congruent
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So in short, the answer is choice B) 27; 49