One proof proves an angle to be 90º, and the other proof proves the very same angle to be 60º. I've looked over both proofs several times but I can't find what the error in either one of them is. And they obviously can't both be correct.
Below are two pictures from my notebook. One states the problem itself and exactly what type of shapes and angles we're dealing with. The first picture also shows the first proof in two column format. The second picture shows the second proof also in two column format.
Can anybody spot a problem with either one of these?
EDIT: Just to be clear I actually don't know what the problem with one of these proofs is. That's why I'm asking. Both seem like they should work to me, yet they yield contradicting conclusions.
Initial Problem
Given the following:
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$ABCD$ is a square
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$\overline{FG}$ is a perpendicular bisector of $\overline{BG}$
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$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AC}$ is part of a circle centered at $B$
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Points $E$, $A$, and $C$ all lie on circle $B$
Find $m\angle BEC$
Proof that $m\angle BEC=60^\circ$
$\begin{aligned}
& &&\text{Statement} &&\text{Reason}\\
&1. &&\overline{BG}≅\overline{GC} &&\text{Given}\\
&2. &&m<CGE=90^\circ &&\text{Given}\\
&3. &&m\angle BGE=90^\circ &&\text{Given}\\
&4. &&\overline{EG}≅\overline{FG} &&\text{Reflexive property}\\
&5. &&\triangle CGE≅\triangle BGE &&\text{SAS}\\
&6. &&\overline{BE}≅\overline{CE} &&\text{Because of $5$}\\
&7. &&\text{Both $\overline{BE}$ and $\overline{BG}$ are radii of circle $B$} &&\text{Both $E$ and $C$ lie on circle $B$ and point $B$ is the origin of circle $B$}\\
&8. &&\therefore \overline{BC}≅\overline{BE} &&\text{The radius of the same circle remains constant}\\
&9. &&\overline{BC}≅ EC &&\text{Transitive property}\\
&10. &&\triangle BEC\text{ is equilateral} &&8\text{ and }9\\
&11. &&m\angle BEC = 60^\circ &&\text{All angles in an equilateral triangle are $60^\circ$}\\
&&&\text{Q.E.D.}
\end{aligned}$
Proof that $m\angle BEC=90^\circ$
$\begin{aligned}
& &&\text{Statement} &&\text{Reason}\\
&1. &&\text{construct line $\overline{HI}$ such that $\overline{HI}⟂\overline{AB}$ and point $E$ is collinear to $\overline{HI}$} &&\text{None}\\
&2. &&m<CGE\text{ and } m\angle BGE\text{ are $90^\circ$} &&\text{Given}\\
&3. &&\overline{CG}≅\overline{BG} &&\text{Given}\\
&4. &&\overline{EG}≅\overline{EG} &&\text{Reflexive property}\\
&5. &&\triangle CGE ≅ \triangle BGE &&\text{SAS}\\
&6. &&\overline{HE}≅\overline{BG} &&\text{Because of how we constructed $\overline{HI}$}\\
&7. &&\overline{HB}≅\overline{EG} &&\text{Same reason as $6$}\\
&8. &&m\angle EGB=90^\circ &&\text{Same reason as $6$}\\
&9. &&\triangle GEB≅\triangle HEB &&\text{SAS}\\
&10. &&m\angle HEB = m\angle GEB &&\text{Statement $9$}\\
&11. &&m\angle HEG = 90^\circ &&\text{Same reason as $6$}\\
&12. &&m\angle HEB + m\angle GEB = 90^\circ &&\text{None}\\
&13. &&2(m\angle HEB)= 90^\circ&&\text{Statement $10$}\\
&14. &&m\angle HEB= 45^\circ&&\text{None}\\
&15. &&m\angle BEG= 45^\circ&&\text{Statement $10$}\\
&16. &&m\angle CEG= 45^\circ&&\text{Statement $5$}\\
& &&\therefore m\angle BEC = m\angle CEG + m\angle BEG = 45^\circ + 45 ^\circ =90^\circ&&\\
&&&\text{Q.E.D.}
\end{aligned}$
Best Answer
You made a mistake in statement (9) of the second exercise. If you apply SAS to $\triangle BEH$ and $\triangle EBG$, you obtain that $\angle BEG=\angle EBH$ and $\angle GBE=\angle BEH$.