In ∆ABC, all the sides are equal (each equals 5). Hence, the triangle is equilateral, and ∠A = ∠B = ∠C = 60°.

Also, ∆EFG is a right-angled isosceles triangle (since EG = FG = 5), and The Pythagorean Theorem is satisfied \((EG^2+ FG^2 = 5^2 + 5^2 = 50 = EF^2 = (5\sqrt{2})^2\)

Hence, ∠E = ∠F = 45° (Angles opposite equal sides of an isosceles right triangle measure 45° each).
Now, in ∆CED, we have:∠D = x° vertical angles, from the figure ∠C = ∠C in ∆ABC vertical angles, from the figure = 60° we know ∠E in ∆CED = ∠E in ∆EFG vertical angles = 45° we know

Now, summing these three angles of ∆CED to 180° yields 60 + 45 + x = 180. Solving this equation for x yields x = 75. The answer is (E).

The OA is correct but the figure was slightly different


I fixed the question

OE