Let BC be a fixed line segment in the plane. | vedclass

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(D) For $	riangle ABC$ to be isosceles,at least two sides must be equal. Let the fixed segment be $BC$ with length $a$.Case $I$: $AB = AC$. The locus

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the union of two circles and a line

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(D) For $	riangle ABC$ to be isosceles,at least two sides must be equal. Let the fixed segment be $BC$ with length $a$.Case $I$: $AB = AC$. The locus of $A$ is the perpendicular bisector of the segment $BC$,which is a straight line.Case $II$: $AB = BC$. Since $BC$ is fixed,$AB$ must be equal to the constant length $BC$. Thus,the locus of $A$ is a circle centered at $B$ with radius equal to $BC$.Case $III$: $AC = BC$. Similarly,the locus of $A$ is a circle centered at $C$ with radius equal to $BC$.Therefore,the complete locus of $A$ is the union of the perpendicular bisector of $BC$ and two circles centered at $B$ and $C$ with radius $BC$ (excluding points where $A, B, C$ are collinear,which are the finitely many exceptional points).Thus,the correct option is $(d)$.

Let $BC$ be a fixed line segment in the plane. The locus of a point $A$ such that the $\triangle ABC$ is isosceles,is (with finitely many possible exceptional points)

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