Question

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

Answer 1

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)`