[tex]\text{If}\ \overline{BD}\ \text{is bisects of}\ \angle ABC,\ \text{then}\ m\angle ABD=m\angle DBC.\\\\\text{We have}\ m\angle ABC=7x,\ \text{and}\ m\angle ABD=3x+25\to m\angle DBC=3x+25\\\\m\angle ABC=m\angle ABD+m\angle DBC\\\\\text{substitute}\\\\7x=(3x+25)+(3x+25)\\\\7x=3x+25+3x+25\qquad|\text{combine like terms}\\\\7x=(3x+3x)+(25+25)\\\\7x=6x+50\qquad|\text{subtract 6x from both sides}\\\\x=50\\\\m\angle DBC=3(50)+25=150+25=175\\\\Answer:\ m\angle DBC=175^o[/tex]
