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# Geometry Question

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http://imgur.com/a/qYGiy

I wasn't able to upload a picture for some reason so I made a url. Thank you for your help. :)

Gwendolynkristine  Jun 6, 2017

#1
+4784
+1

$$\angle BCD=\frac12(\overset{\frown}{AE}-\overset{\frown}{BD}) \\~\\ 36^{\circ}=\frac12(\overset{\frown}{AE}-26^{\circ}) \qquad\text{ Multiply both sides of the equation by 2.} \\~\\ 2\,*\,36^{\circ}=\overset{\frown}{AE}-26^{\circ}\qquad\text{Add 26 degrees to both sides of the equation.}\\~\\ 2\,*\,36^{\circ}+26^{\circ}=\overset{\frown}{AE} \\~\\ 98^{\circ}=\overset{\frown}{AE}$$

hectictar  Jun 6, 2017
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#1
+4784
+1

$$\angle BCD=\frac12(\overset{\frown}{AE}-\overset{\frown}{BD}) \\~\\ 36^{\circ}=\frac12(\overset{\frown}{AE}-26^{\circ}) \qquad\text{ Multiply both sides of the equation by 2.} \\~\\ 2\,*\,36^{\circ}=\overset{\frown}{AE}-26^{\circ}\qquad\text{Add 26 degrees to both sides of the equation.}\\~\\ 2\,*\,36^{\circ}+26^{\circ}=\overset{\frown}{AE} \\~\\ 98^{\circ}=\overset{\frown}{AE}$$

hectictar  Jun 6, 2017

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