Math  /  Geometry

Question6. In the figure below, ray EFundefined\overrightarrow{E F} was constructed starting from rays EDundefined\overrightarrow{E D} and ECundefined\frac{\overrightarrow{E C}}{}. By using a compass DD and GG were marked equidistant from EE on rays EDundefined\overrightarrow{E D} and EGundefined\overrightarrow{E G}. The compass was then used to locate a point FF, distinct from EE, so that FF is equidistant from DD and GG. For all constructions defined by the above steps, the measures of DEF\angle D E F and GEF\angle G E F :

Studdy Solution
To conclude, DEF=GEF\angle DEF = \angle GEF because EFundefined\overrightarrow{EF} is the angle bisector of DEG\angle DEG.
Therefore, DEF\angle DEF and GEF\angle GEF are equal, and since EFundefined\overrightarrow{EF} bisects DEG\angle DEG, each measure is 12DEG\frac{1}{2} \angle DEG.

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