[Solutions Posted]Geometry Proof Problem of the day #3 (Diagram Added)

Hey I got around more than half way to the problem. It's almost completed. Feeling tired. I will post it tommorow. (Feeling very excited but Tired). :P :P

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Hurray!!! I got the solution.

Since $EABC$ is cyclic. It implies that $\angle ABC = 180^{0} - \angle AEC = \angle CEG$.

We have to prove that $CE = EG$ i.e. $\angle CEG = \angle CGE$. It suffices to prove that $\angle CGE = \angle ABD$.

Let '$I$' be foot of perpendicular from '$C$' onto $EG$. But, $\angle EFC = \angle EIC = 90^{0}$. So, $\Delta BFC \sim \Delta CEI$. So, $\dfrac{BF}{BC} = \dfrac{EI}{EC}$ (#)

Observe that $\angle DBF = 180^{0} - \angle DEA = \angle DEG$ (1)

Since $ED$ is diameter of the circle. It implies that $\angle DAE = 90^{0}$. But $\angle DFG = 90^{0} = \angle DAG$. So, $A, G, D, F$ are cyclic.

It implies that $\angle EGD = 180^{0} - \angle AFD = \angle BFD$ (2)

From (1) and (2) it implies that $\Delta BFD \sim \Delta EGD$. So, $\dfrac{BF}{BD} = \dfrac{EG}{ED}$(##)

Dividing (#) by (##), we get,

$\dfrac{BD}{BC} = \dfrac{EI}{EC} \times \dfrac{ED}{EG}$ $\dfrac{EI}{EG} = \dfrac{BD}{BC} \times \dfrac{EC}{ED} = \dfrac{1}{2} \times \dfrac{BD}{BN} \times \dfrac{EC}{ED}$

Since, $\Delta ECD$ is right angled at '$C$' and also $\angle CED = \angle CBD$. So, $\cos{\angle CED} = \cos{\angle CBD}$. It implies that $\dfrac{EC}{ED} = \dfrac{BN}{BD}$

It gives out that $\dfrac{EI}{EG} = \dfrac{1}{2}$ i.e. $EG = 2 EI$. So, $EI =GI$. It implies that $\angle CEG = \angle CGE$ i.e. $CG = CE$ as required.

SPECIAL THANKS TO @Xuming Liang

A slightly alternate solution:

Since $DA\perp AG, DF\perp GF$, $D,F,A,G$ are concyclic and thus $\angle DGA=\angle DFB$. In addition,note that $\angle DBA=\angle DEG$, therefore $DEG\sim DBF$. At this point one can use the new-found information to ratio chase $GE$ related lengths, as Surya Prakash has demonstrated, or we can continue to utilize similar (they are almost the same thing),

Since $DEG\sim DBF$, we ask what point corresponds to $C$ in the $DBF$ configuration, i.e. For what point is it to $DBF$ as $C$ is to $DEG$(Not sure if this makes sense...) By some thought we quickly see that it is the midpoint of $BC$ denoted by $M$ (I will elaberate: it must lie on $BC$ since $\angle GEC=\angle FBC$, and $M$ is the only point that satisfy $BM=FM$). Clearly $DBM\sim DEC$, thus $DEGC\sim DBFM$. SInce $BM=FM$, we have $CG=EC$ and are done.

Similarity is powerful isn't it?

Can you post the figure?

So basically, we have to prove that G and E coincide, right?

@Xuming Liang

@Xuming Liang Do D and E lie on the circumcirlce of $\Delta ABC$?