In the figure alongside , O is the center of the circle . Chord AB is parallel to the chord CD and BC is a
diameter. Proved that arc AC = arc BD.
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We know that ∠BCD and ∠ABC are alternate interior angles
∠BCD=∠ABC=25
o
We know that the angle subtended by an arc of a circle at the center is double the angle subtended by the arc at any point on the circumference
∠BOD=2∠BCD
It is given that ∠BCD=25
o
So we get
∠BOD=2(25
o
)
By multiplication
∠BOD=50
o
In the same way
∠AOC=2∠ABC
So we get
∠AOC=50
o
From the figure, we know that AB is a straight line passing through the centre
Using the angle sum property
∠AOC+∠COD+∠BOD=180
o
By substituting the values
50
o
+∠COD+50
o
=180
o
On further calculation
∠COD+100
o
=180
o
By subtraction
∠COD=180
o
−100
o
so we get
∠COD=48
o
We know that
∠CED=
2
1
∠COD
So we get
∠CED=
2
80
o
By division
∠CED=40
o
Therefore, ∠CED=40
o
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