I am quite new to inversion so I tried this, please let me know if there are any errors $\triangle CAD \sim \triangle CEA \implies CA^2 = CD \times CE$. Invert about $C$ with radius $CA$, let $E'$ be the inverse of $E.$ By definition $CE \times CE' = CA^2 = CE \times CD \implies E' \equiv D$ similarly if $F'$ is the inverse of $F$ then $F' \equiv G \implies CE \times CD = CF \times CG \implies \dfrac{CD}{CF} = \dfrac{CG}{CE}$. Using the inversion distance formula we get that $E' F' = \dfrac{CA^2 \times EF}{CE \times CF} \implies DG = \dfrac{CD \times EF}{CF} = EF \times \dfrac{CD}{CF} = EF \times \dfrac{CG}{CE}$ $$\implies CE \times DG = EF \times CG.$$

Note that $\measuredangle CEF=\measuredangle ABF+\measuredangle BFC=\measuredangle DGC\implies \triangle CEF\sim \triangle CGD,$ which yields conclusion.

MODS PoTD Claim 1: $D,E,F,G$ are cyclic. Proof: Note that \[\angle CEF = \angle CAB - \angle FCB = 45 - \angle GCB = \angle CGB = \angle FGD\]$\square$ Then, we clearly have $\triangle CFE \sim \triangle CDG$ since $C,E,D$ and $C,F,G$ are collinear. Thus, we have \[\frac{CE}{EF} = \frac{CG}{DG} \Longrightarrow CE\cdot DG = EF\cdot CG\]and we're done. $\blacksquare$.

$$\angle DGC = \angle BCG + \angle GBC = \angle BEF + \angle CAB = \angle BEF + \angle CEB = \angle CEF$$so $\triangle CDG \sim \triangle CFE$. $\frac{CG}{DG} = \frac {CE}{EF}$ which is equivalent to the desired result.

parmenides51 wrote: An isosceles triangle $ABC$ is inscribed in a circle with $\angle ACB = 90^o$ and $EF$ is a chord of the circle such that neither E nor $F$ coincide with $C$. Lines $CE$ and $CF$ meet $AB$ at $D$ and $G$ respectively. Prove that $|CE|\cdot |DG| = |EF| \cdot |CG|$. Wasn't it enough for the triangle to be isosceles? Best regards, sunken rock