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,

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

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}$