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BD bisects ZABC.Find mZABD, mZCBD, and m ZABC.AA (6x + 14)D(3x 29)BC

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3 votes

We know that


\vec{BD}\text{ bisects }\angle ABC

That means


m\angle ABD=m\angle CBD

So,


\begin{gathered} (6x+14)^(\circ)=(3x+29)^(\circ) \\ 6x+14=3x+29 \\ 6x-3x=29-14 \\ 3x=15 \\ x=(15)/(3)=5 \end{gathered}

If


\begin{gathered} \vec{BD}\text{ bisects }\angle ABC \\ \text{then} \\ m\angle ABC=m\angle ABD+m\angle CBD \end{gathered}

Finally,


m\angle ABD=(6x+14)^(\circ)=(6\cdot5+14)^(\circ)=44^(\circ)
m\angle CBD=(3x+29)^(\circ)=(3\cdot5+29)^(\circ)=44^(\circ)
m\angle ABC=44^(\circ)+44^(\circ)=88^(\circ)

User Inzimam Tariq IT
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