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# The angle bisector of an angle divides the angle into two angles with equal measure. In the diagram below, $A$ is on segment $\overline{CE}$

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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?

Guest Oct 26, 2017

#2
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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...

hectictar  Oct 26, 2017
edited by hectictar  Oct 26, 2017
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#1
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I don't know that...

Conman13  Oct 26, 2017
#2
+5254
+3

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...