Given circles \( D \) and \( E \), with \( \overleftrightarrow{G J I} \) tangent to \( D \) at point \( G \) and to \( E \) at point \( I \) and \( \overleftarrow{F J H} \) tangent to \( D \) at point \( F \) and to \( E \) at point \( H \). Identify a segment that is congruent to \( \overline{F J} \).

To determine a segment that is congruent to \( \overline{F J} \), we can utilize the properties of tangents drawn from an external point to a circle. Since \( J \) is a point outside both circles \( D \) and \( E \), and \( \overleftrightarrow{G J I} \) is tangent to \( D \) at point \( G \) and to \( E \) at point \( I \), we know that the segments \( \overline{G J} \) and \( \overline{I J} \) are both tangent to their respective circles from point \( J \). In addition, the segment \( \overleftarrow{F J H} \) is tangent to circle \( D \) at point \( F \) and to circle \( E \) at point \( H \). This means \( \overline{F J} \) and \( \overline{H J} \) are also tangents from point \( J \). By the tangent segment theorem, we know that the lengths of tangents from a common external point are equal. Thus, we have: \[ \overline{F J} \cong \overline{G J} \] and \[ \overline{H J} \cong \overline{I J}. \] Therefore, the segment \( \overline{G J} \) is congruent to \( \overline{F J} \).