Given: See the diagram.Prove: DC = DBStatementReason1. given2. AG = GC given3. is the perpendicular bisector of . deduced from steps 1 and 24. DA = DC 5. given6. AH = HB given7. is the perpendicular bisector of . definition of perpendicular bisector8. DA = DB deduced from steps 6 and 79. DC = DB Transitive Property of EqualityWhat is the reason for the fourth and eighth steps in the proof?

Answer:A. ASA criterion for congruent triangles.Step-by-step explanation:Given DG is perpendicular bisector of CA and DH is perpendicular bisector of AB.In triangle DGC and DGADG=DG( reflexive property of equality)[tex]\angle DGA=\angle DGC=90^{\circ}[/tex] ( given )[tex]\angle ADG=\angle CDG[/tex] ( by definition of perpendicular bisector)[tex]\therefore \triangle AGD\cong \triangle CGD[/tex] ( ASA postulate) Similarly, In triangle ADH and triangle BDHDH=DH ( reflexive property of equality)[tex]\angle DHA=\angle DHB=90^{\circ}[/tex] (given)[tex]\angle ADH=\angle BDH[/tex] ( By definition of perpendicular bisector)[tex]\therefore \triangle ADH\cong \triangle BDH[/tex] ( ASA postulate)1.Statement: [tex]\overline{DG}\perp \overline{AC}[/tex]Reason: Given.2. Statemnet: AG=GCReason: Given 3. Statement: [tex]\overline{DG}[/tex] is perpendicular bisector of [tex]\overline{AC}[/tex]Reason: from step 1 and step 2.4.Statement: DA=DCReason: ASA criterion for congruent triangles.5 .Statement:[tex]\overline{DH} \perp \overline{AB}[/tex]Reason: Given6.  Statement:AH=HBReason:Given7.Statement: [tex]\overline{DH}[/tex] si perpendicular bisector [tex]\overline{AB}[/tex]Reason: By definition of perpendicular bisector.8.Statement: DA=DBReason : ASA criterion for congruent triangles.9.Statement: DC=DBReason: Transitive property of equality.Hence proved.