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Chord Confluence


In a cyclic quadrilateral $ABCD$, the diagonals intersect at $E$. $F$ and $G$ are on chord $AC$ and chord $BD$ respectively such that $AF$ = $BE$ and $DG$ = $CE$. Prove that, $B,G,F,C$ lie on the same circle.


Source: BdMO 2024 National Secondary P2


Proof Based Problems  


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Solution

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$AC$ and $BD$ intersect at E. So we use the power of point. 


We have,

$BE.DE = CE.AE$

$\Rightarrow \frac{DE}{CE} = \frac{AE}{BE} $

$ \Rightarrow \frac{DG + GE}{CE}=\frac{AF + FE}{BE}$

$\Rightarrow \frac {CE + GE}{CE}=\frac{BE + FE}{BE}$

$\Rightarrow 1 + \frac{GE}{CE}= 1 + \frac{FE}{BE}$

$\Rightarrow \frac{GE}{CE}= \frac{FE}{BE}$

$\Rightarrow GE.BE = FE.CE$


So, by power of point we have that $BFGC$ has to be cyclic.

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