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
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.
To prove that FEHG is a cyclic quadrilateral, we need to show that angles ∠FEG and ∠FHG are supplementary.
Given that circle BCGH passes through points B and C, we can say that ∠BGH = ∠BCH and ∠CGH = ∠CBH. This is because angles inscribed in the same arc (in this case, arc BH) are equal.
Since BCEF is a parallelogram, we know that ∠BCF = ∠BEC. Also, since BC is a straight line, we know that ∠BCH + ∠CBH = 180°. Thus, we can rewrite these angles as ∠BCF = ∠BEC = 180° - ∠CGH = 180° - ∠CBH.
Now, let's consider the quadrilateral FEHG.
Since BCEF is a parallelogram, we know that ∠BEC = ∠FEC and ∠BCF = ∠FBC.
Using the property of alternate angles, we can say that ∠FEC = ∠GHC and ∠FBC = ∠FGH.
Combining these results, we have:
∠FEG = ∠FEC - ∠GEC
= ∠BEC - ∠GEC
= ∠BCF - ∠GEC
= ∠BCF - ∠CGH
= ∠BCF - (180° - ∠CBH)
= ∠BCF - 180° + ∠CBH
= ∠FBC - 180° + ∠CBH
= ∠FGH - 180° + ∠CBH
= ∠FHG + ∠CBH - 180°
Since ∠FHG is the supplement of ∠FEG, we can conclude that FEHG is a cyclic quadrilateral.