Question

gj bisects ∠fgh and is a perpendicular bisector of fh. what is true of triangle fgh?
options:
○ it is a right triangle.
○ it is an obtuse triangle.
○ it has exactly 2 congruent sides.
○ it has exactly 3 congruent sides.
image: triangle fgh with g at top, f and h at base, j on fh (midpoint) with gj perpendicular to fh (right angle symbol) and gj bisecting ∠fgh (30° angle on each side of gj)

Explanation:

1. Since \( \overline{GJ} \) is the perpendicular bisector of \( \overline{FH} \), \( FJ = JH \) and \( \angle GJF=\angle GJH = 90^\circ \). Also, \( \overline{GJ} \) is common to triangles \( \triangle GJF \) and \( \triangle GJH \). By SAS (Side - Angle - Side) congruence criterion, \( \triangle GJF\cong\triangle GJH \). So, \( GF = GH \), which means two sides are congruent.
2. Given that \( \overline{GJ} \) bisects \( \angle FGH \) and one of the angles formed is \( 30^\circ \), then \( \angle FGH=60^\circ \). Since \( GF = GH \) and \( \angle FGH = 60^\circ \), by the property of isosceles triangles (if an isosceles triangle has a \( 60^\circ \) angle, it is equilateral), all three sides \( GF = GH=FH \). So the triangle has exactly 3 congruent sides.

• For the first option, a right triangle has one \( 90^\circ \) angle. The angles in our triangle: since it's equilateral, all angles are \( 60^\circ \), so it's not a right triangle.
• For the second option, an obtuse triangle has one angle greater than \( 90^\circ \). Our triangle has all angles \( 60^\circ \), so it's not obtuse.
• For the third option, we just proved it has 3 congruent sides, not exactly 2.

Answer:

It has exactly 3 congruent sides.