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see the diagram.
We need to prove that sum of the angles made by the two arcs on the LHS or RHS is 180°. Let the chords AB and CD intersect at E.
In the right angle ΔDEB, ∠EDB + ∠EBD = 90°
∠COB = 2 * ∠CDB (an arc makes at center twice the angle it makes on the circle.
∠AOD = 2 * ∠ABD
∠COB + ∠AOD = 2 (∠CDB + ∠ABD) = 2*90°=180°
Since total angle made by the arcs AXD and CZB is 180°, their total length is half the perimeter of the circle. So arcs AXD+ arc CZB is a semicircle.
We need to prove that sum of the angles made by the two arcs on the LHS or RHS is 180°. Let the chords AB and CD intersect at E.
In the right angle ΔDEB, ∠EDB + ∠EBD = 90°
∠COB = 2 * ∠CDB (an arc makes at center twice the angle it makes on the circle.
∠AOD = 2 * ∠ABD
∠COB + ∠AOD = 2 (∠CDB + ∠ABD) = 2*90°=180°
Since total angle made by the arcs AXD and CZB is 180°, their total length is half the perimeter of the circle. So arcs AXD+ arc CZB is a semicircle.
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