If △EFG ≅ △HJK, complete each part.

Answer:

[tex] \overline{EG} \cong \overline{HK} [/tex]

[tex] \overline{HJ} \cong \overline{EF} [/tex]

[tex] \angle F \cong \angle J [/tex]

[tex] \angle H \cong \angle E [/tex]

∆FGE ≅ ∆JKH

Step-by-step explanation:

Since ∆EFG ≅ ∆HJK, therefore, the corresponding sides and angles of both ∆s are congruent.

Thus:

[tex] \overline{EG} \cong \overline{HK} [/tex]

[tex] \overline{HJ} \cong \overline{EF} [/tex]

[tex] \angle F \cong \angle J [/tex]

[tex] \angle H \cong \angle E [/tex]

∆FGE ≅ ∆JKH (take note of the arrangement of the letters or naming of each vertex of both triangles. Each vertex of in one ∆ must correspond to the vertex of the other for the ∆s to be congruent)

△EFG ≅ △HJK,

From the figure attached we can conclude  that

  [tex]\rm \bar {EG}[/tex] ≅ [tex]\rm \bar {HK}[/tex]

  [tex]\rm \bar HJ[/tex] ≅ [tex]\rm \bar FE[/tex]

   ∠[tex]\rm F[/tex] ≅∠[tex]\rm J[/tex]

   ∠[tex]\rm H[/tex] ≅∠[tex]\rm E[/tex]

△FGE ≅ △JKH

Given that

△EFG ≅ △HJK,

Since △EFG ≅ △HJK is given hence we can conclude that corresponding sides and corresponding angles of both the triangles will be equal.

Two triangles are said to be congruent when they are of similar shape and size means all the angles of one triangle are  equal to the corresponding angles of other triangle and  all the  sides of one triangle are equal to  corresponding sides of other triangle.

From the figure attached we can conclude  that

  [tex]\rm \bar {EG}[/tex] ≅ [tex]\rm \bar {HK}[/tex]

  [tex]\rm \bar HJ[/tex] ≅ [tex]\rm \bar FE[/tex]

   ∠[tex]\rm F[/tex] ≅∠[tex]\rm J[/tex]

   ∠[tex]\rm H[/tex] ≅∠[tex]\rm E[/tex]

△FGE ≅ △JKH

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