Question

9 $overrightarrow{bd}$ bisects $angle abc$ in the figure shown. $x = square$ $mangle abc = square^circ$ (figure shows $angle abd = 70^circ$, $angle dbc = 10x^circ$ with vertex at $b$, sides $ba$, $bd$, $bc$)

Explanation:

Step1: Use angle bisector definition

Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), \(\angle ABD=\angle CBD\). So \(10x = 70\).

Step2: Solve for \(x\)

Divide both sides by 10: \(x=\frac{70}{10}=7\).

Step3: Calculate \(m\angle ABC\)

\(m\angle ABC=\angle ABD+\angle CBD = 70 + 70=140\) (or using \(10x+70\), substitute \(x = 7\): \(10\times7 + 70=70 + 70 = 140\)).

Answer:

\(x = 7\), \(m\angle ABC=140\)