Step 1: Set up the triangle and the known angle.
Let the base be AB = 60 cm, with the base angle at A equal to \(60^{\circ}\). Call BC = a and AC = b, so a and b are the two sides other than the base. We want the shortest of a, b and 60.
Step 2: Test statement I alone.
Statement I says \(a + b = 80\). Using the cosine rule at angle A:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad c = 60 \]
Since \(\cos 60^{\circ} = \frac{1}{2}\):
\[ \frac{1}{2} = \frac{b^2 + 3600 - a^2}{120b} \implies 60b = 3600 + b^2 - a^2 \]
Replace \(a = 80 - b\) (from statement I), so \(a^2 = 6400 - 160b + b^2\):
\[ 60b = 3600 + b^2 - (6400 - 160b + b^2) = 160b - 2800 \] \[ 100b = 2800 \implies b = 28, \quad a = 80 - 28 = 52 \]
This pins down every side, 60, 52 and 28, and it checks out with the cosine rule directly (\(52^2 = 28^2 + 60^2 - 2(28)(60)(0.5)\), which works out to 2704 = 2704). So statement I alone fixes the triangle completely, and the shortest side is 28 cm. Statement I is sufficient by itself.
Step 3: Test statement II alone.
Statement II says the other base angle, B, is \(45^{\circ}\). Since the angles of a triangle add to \(180^{\circ}\):
\[ C = 180^{\circ} - 60^{\circ} - 45^{\circ} = 75^{\circ} \]
Now all three angles are known, and one side, the base of 60 cm, is known, so the sine rule fixes the other two sides uniquely:
\[ \frac{a}{\sin 60^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{60}{\sin 75^{\circ}} \]
Every ratio here is a known number, so a and b can each be calculated. A triangle with one side and all three angles known is completely fixed, so statement II alone is also sufficient.
Step 4: Compare the two.
Both statements independently pin down a complete, unique triangle. They describe two different triangles, since the extra condition differs, but that does not matter for data sufficiency, since each condition on its own answers the question. So either statement I alone or statement II alone answers the question.
Final Answer:
Either statement alone is sufficient, which is option (DD).