easy using sin law. it's just a few lines of writing...

Can you post your solution? or at least the idea?

Suppose, $ \angle DEG=\alpha $ and $ \angle CEG=\beta $ So $ \frac {DG}{CG}=\frac {DE.\sin {\alpha}}{SC.\sin {\beta}} $ and similar for the other. Comparing those two equations we get $ \frac {\sin^2 {\alpha}}{\sin^2 {\beta}}=\frac {EC.AE}{EB.DE} $ So $ \frac {DG^2}{CG^2}=\frac {AE.DE}{EB.EC}\implies \frac {AD^2}{BC^2}=\frac {AE.DE}{EB.EC}=\frac {[AED]}{[BEC]} $ Now make a homothety $ (E,-\frac {BC}{AD}) $ and suppose the images of $ A,D $ under this homothety are $ A',D' $. Note that $ [EA'D']=[EBC] $. So $ BA'\parallel CD' $ and $ BC=A'D' $. So $ BA'CD' $ is a isosceles trapezium. So $ B,C,A',D' $ are cyclic. So $ B,C,A,D $ are cyclic.