The angle bisector of an angle divides the angle into two angles with equal measure. In the diagram below, \(A\) is on segment \(\overline{CE}\) and \(\overline{AB}\) bisects \(\angle CAD\). If we have \(\overline{DA}\parallel\overline{EF}\) and \(\angle AEF\) is \(12^\circ\) less than 10 times \(\angle BAC\), then what is \(\angle CAD\) in degrees?
+9481 
If we look at it like this, we can see that ∠AEF and ∠CAD will add to 180° .
∠AEF + ∠CAD = 180° Plug in 10( ∠BAC ) - 12° for ∠AEF .
10( ∠BAC ) - 12° + ∠CAD = 180° And since AB bisects ∠CAD, ∠CAD = 2( ∠BAC ) .
10( ∠BAC ) - 12° + 2( ∠BAC ) = 180° Add 12° to both sides of this equation.
10( ∠BAC ) + 2( ∠BAC ) = 180° + 12° Combine like terms.
12( ∠BAC ) = 192° Divide both sides by 12 .
∠BAC = 16°
And remember that
∠CAD = 2( ∠BAC ) So...
∠CAD = 2( 16° )
∠CAD = 32°
+9481
If we look at it like this, we can see that ∠AEF and ∠CAD will add to 180° .
∠AEF + ∠CAD = 180° Plug in 10( ∠BAC ) - 12° for ∠AEF .
10( ∠BAC ) - 12° + ∠CAD = 180° And since AB bisects ∠CAD, ∠CAD = 2( ∠BAC ) .
10( ∠BAC ) - 12° + 2( ∠BAC ) = 180° Add 12° to both sides of this equation.
10( ∠BAC ) + 2( ∠BAC ) = 180° + 12° Combine like terms.
12( ∠BAC ) = 192° Divide both sides by 12 .
∠BAC = 16°
And remember that
∠CAD = 2( ∠BAC ) So...
∠CAD = 2( 16° )
∠CAD = 32°
