yes

This is incomplete question,I can prove if given that chord AB and CD are equal.

even i can do that..

we do not need any such thing.

yes yes

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yes

This is incomplete question,I can prove if given that chord AB and CD are equal.

even i can do that..

we do not need any such thing.

yes yes

Log in to add a comment

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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

==============================================

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

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

==============================================

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

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