1. First, the angle BHC= 135 degrees and the angle AFC=90 degrees, so the angle FBH=45 degrees.

The angle ECH=45 degrees, so the angle a is equal to 45 degrees.

Then the angle BHC is equal to 135 degrees, and because B.G.C is definitely a circle and the angle BHC is equal to BGC, then h must fall on the circle (the same arc corresponds to the same angle), so four points are * * circles.

Let GH extend and intersect with AB at p, then the angle FHP= angle GHC, and the angle GHC= angle GBC=22.5 degrees (the fillets of the same arc are equal).

So I have to prove it. (It's really hard)

2. Root number 2 (very simple, don't you need to prove it? )

I can't see clearly. . C is the focus of the arc AC on the semicircle.