Let \(\Delta ABC\) be an arbitrary triangle.

Point \(P\) is the midpoint of line \(BC\), and line \(PQ\) is perpendicular to line \(BC\).

Line \(AQ\) bisects angle \( \angle BAC\).

Lines \(AB\) and \(AC\) are extended as necessary to include points \(R\) and \(S\) such that \(\Delta AQR\) and \(\Delta AQS\) are right triangles.

By reason of symmetry, line segments \(BQ\) and \(CQ\) are equal.

By reason of symmetry, line segments \(RQ\) and \(SQ\) are equal.

Therefore, by reason of congruent right triangles, line segments \(BR\) and \(CS\) are equal.

But, also by reason of congruent right triangles, line segments \(AR\) and \(AS\) are also equal.

Therefore, line segments \(AB\) and \(AC\) are equal, and triangle \(\Delta ABC\) is isosceles.

Which of the following multiple choice answers is correct?

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