Given: FH ⊥ GH; KJ ⊥ GJ Prove: ΔFHG ~ ΔKJG Identify the steps that complete the proof. ♣ = ♦ = ♠ =

Right angles are congruent therefore, [tex]\rm \angle FHG \cong \angle GJK[/tex]. [tex]\rm \angle FGH\;and \; \angle KGJ[/tex] are vertical angles hence they are also congruent to each other. And according to AA similarity theorem, two triangles are similar if there corresponding angles are congruent, therefore, [tex]\rm \Delta FHG \sim \Delta KJG[/tex].Given :[tex]\rm FH \perp GH[/tex][tex]\rm KJ \perp GJ[/tex]According to the definition of perpendicular lines, [tex]\rm \angle FHG[/tex] and [tex]\rm \angle GJK[/tex] are the right angles.[tex]\rm \angle FHG \cong \angle GJK[/tex] because all right angles are congruent.[tex]\rm \angle FGH\;and \; \angle KGJ[/tex] are vertical angles therefore, angle [tex]\rm \angle FGH[/tex] are congruent to angle [tex]\rm \angle KGJ[/tex].According to AA similarity theorem, [tex]\rm \Delta FHG \sim \Delta KJG[/tex].Right angles are congruent therefore, [tex]\rm \angle FHG \cong \angle GJK[/tex]. [tex]\rm \angle FGH\;and \; \angle KGJ[/tex] are vertical angles hence they are also congruent to each other. And according to AA similarity theorem, two triangles are congruent if there corresponding angles are congruent, therefore, [tex]\rm \Delta FHG \sim \Delta KJG[/tex].For more information, refer the link given below: