A circle through B and C cuts sides BF and CE of parallelogram BCEF respectively at H and G.Prove that FEHG is a cyclic quadrilateral

Answer 1

First, note that since BCEF is a parallelogram, we have angle BFC = angle CEF. Also, since the circle passes through B and C, we have angle BHC = angle BFC and angle CGE = angle CEF. Combining these equalities, we get angle BHC = angle CGE.

Now, let's consider the quadrilateral FEHG. We want to prove that it is cyclic, which means we want to show that angle FHE + angle GFE = 180 degrees. Since BCEF is a parallelogram, we have angle CEF = angle BFC = angle HBC and angle BFC = angle CEF = angle GCE. Using the fact that angle BHC = angle CGE, we can write:

angle FHE + angle GFE
= (180 - angle HBC - angle CEF) + (180 - angle GCE - angle CEF)
= 360 - (angle HBC + angle GCE + 2*angle CEF)
= 360 - 2*angle BHC
= 180

Therefore, angle FHE + angle GFE = 180 degrees, which means that FEHG is indeed a cyclic quadrilateral.