# In figure, AB and

Given AB and CD are two chords of the circle with centre O which intersects at E.

We have to prove that AEC = 1/2 (angle subtended by arc CXA at centre + angle subtended by arc DYB at centre).

Construction:

Join AC, BC and BD.

Proof:

AC is a chord.

We know that angle subtended at center is double the angle subtended at circumference.

⇒∠AOC = 2ABC … (1)

⇒∠DOB = 2DCB … (2)

⇒∠AOC + DOB = 2(ABC + DCB) … (3)

Consider ΔCEB,

We know that the sum of two opposite interior angles is the exterior angle.

⇒∠AEC = ECB + CBE

⇒∠AEC = DCB + ABC … (4)

From (3) and (4),

AOC + DOB = 2 (AEC)

Hence proved.

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