Question

Prove that angles : AOD - BOC=2×BXC
fig attached...

Answer 1

see diagram. there are two of them. The diagram can be symmetric or asymmetric. in both cases, the proof is same.  The proof can be done in two ways.

Mark the angles as shown in figure.  As OA, OB, OC and OD are radii and are equal, OAD, OAB, OBC, OCD form isosceles triangles. So the two angles at the base are same like x, y, z and ω. 

         In OAD,  θ = 180 - 2y 
In triangle BOX, x = exterior angle = sum of angle BOX + angle BXO
In triangle COX,  z = exterior angle = sum of angle COX + angle CXO

        So x + z = (BOX + COX) + (BXO + CXO) = α + β

Now in triangle AXD,  
        β = 180 - (x+y) - (y+z)  = 180 - (x+z) - 2y = θ - α - β

 So   θ = α + 2 β     or    

 AOD - BOC = 2 BXC

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Alternately,

in quadrilateral ABCD,   2 y + 2 x + 2 ω + 2 z = 2π
 so                  x + y + z + ω = π
                    
in AOD,         θ = π - 2 y
in BOC,        α = π - 2 ω 
in XAD,       β = π - 2 y - x - z  =  θ - x - z 

R HS = α + 2 β = π - 2 ω + 2 θ - 2 x  - 2 z
          = 2 θ - π + 2 (π - ω - x - z ) = 2 θ - π + (2 y)
          = 2 θ - (π - 2 y)   =  2 θ - θ  =  θ 
          L H S

Answer 2

see the attachment for explanation soln.------>