Question

question 3 6.25 pts
study this pattern for a quilt.
the quilt panel is a regular hexagon. points h, i,
j, k, l, and m are the midpoints of
\\( \overline { f g } , \overline { a g } , \overline { b g } , \overline { c g } , \overline { d g } \\), and \\( \overline { e g } \\),
respectively.
what facts would be sufficient to prove that the
shaded triangles are all congruent?
select all that apply.
point g is equidistant from points a, b, c, d, e,
and f.
\\( \overline { a g } , \overline { b g } , \overline { c g } , \overline { d g } , \overline { e g } \\), and \\( \overline { f g } \\) are all
congruent.
\\( \delta a g b , \delta b g c , c g d , \delta d g e , \delta e g f \\),
and \\( \delta f g a \\) are all equilateral.
\\( \angle a b c , \angle b c d , \angle c d e , \angle d e f , \angle e f a \\),
and \\( \angle f a b \\) are all congruent.

Explanation:

• For the first option:
• In a regular hexagon, the center \(G\) is equidistant from all the vertices \(A,B,C,D,E,F\). If \(G\) is equidistant from \(A,B,C,D,E,F\) (\(GA = GB=GC = GD=GE = GF\)), and \(H,I,J,K,L,M\) are mid - points (so \(HI=\frac{1}{2}AB\), \(IJ = \frac{1}{2}BC\) etc. and \(AB = BC=CD=DE=EF = FA\) in a regular hexagon). Using the Side - Side - Side (SSS) congruence criterion (if we consider the sides of the shaded triangles formed by the mid - points and the center), the shaded triangles can be proved congruent.
• For the second option:
• If \(\overline{AG},\overline{BG},\overline{CG},\overline{DG},\overline{EG},\overline{FG}\) are all congruent (\(AG = BG=CG = DG=EG = FG\)), and since \(H,I,J,K,L,M\) are mid - points (\(AH=\frac{1}{2}AG\), \(BI=\frac{1}{2}BG\) etc. so \(AH = BI=CK=DL=EM=FH\)), and the angles between the segments (derived from the symmetry of the hexagon) are equal. By SSS (if we consider the sides of the shaded triangles), the shaded triangles are congruent.
• For the third option:
• If \(\triangle AGB,\triangle BGC,\triangle CGD,\triangle DGE,\triangle EGF,\triangle FGA\) are all equilateral. Then \(AG = BG=CG = DG=EG = FG=AB = BC=CD=DE=EF = FA\). Since \(H,I,J,K,L,M\) are mid - points, the sides of the shaded triangles (e.g., if we consider two shaded triangles, the lengths of their sides can be shown to be equal using the mid - point property and the equality of the sides of the equilateral triangles). By SSS, the shaded triangles are congruent.
• For the fourth option:
• The angles \(\angle ABC,\angle BCD,\angle CDE,\angle DEF,\angle EFA,\angle FAB\) being congruent only gives information about the angles of the hexagon. It does not directly provide information about the sides of the shaded triangles. We need information about the sides (either directly or through the properties of equilateral triangles or equal - distance from a point) to prove triangle congruence.

Answer:

• Point \(G\) is equidistant from points \(A,B,C,D,E,\) and \(F\).
• \(\overline{AG},\overline{BG},\overline{CG},\overline{DG},\overline{EG},\) and \(\overline{FG}\) are all congruent.
• \(\triangle AGB,\triangle BGC,\triangle CGD,\triangle DGE,\triangle EGF,\) and \(\triangle FGA\) are all equilateral.