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Question
ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = `1/2` ∠BAD
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Solution
In a circle, ABCD is a quadrilateral having center A.
To prove ∠CBD + ∠CDB = `1/2` ∠BAD
Construction: Join AC.
Proof: As we know that angle subtended by an arc at the center is double the angle subtended by it at point on the remaining part of the circle.
So, ∠CAD = 2∠CBD ...(i)
And ∠BAC = 2∠CDB ...(ii)
Now, adding equation (i) and (ii), we get
∠CAD + ∠BAC = 2(∠CBD + ∠CDB)
∠BAD = 2(∠CBD + ∠CDB)
Hence, ∠CBD + ∠CDB = `1/2` ∠BAD.
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