Connecting DE

* ADB =∠AEB=90 degrees

∴A, B, D, E four-point * * * cycle

So ∠CED=∠ABC

And l is the tangent of the intersection c of ≥ o.

∴∠ACG=∠ABC

∴∠CED=∠ACG

So de FG

So DF=EG

That's right. I was wrong.