Q. 63.7( 6 Votes )

# ABCD is such a quadrilateral that A is the center of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = ∠BAD.

Answer :

Join CA and BD.

Since arc DC subtends ∠DAC at the center and ∠CAB at point B in the remaining part of the circle, we have:

∠DAC = 2∠CBD ……………(1)

(Reason: In a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.)

Similarly, arc BC subtends ∠CAB at the center and ∠CDB at point D in the remaining part of the circle, we have:

∠CAB = 2∠CDB ……………(2)

(Reason: In a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.)

From equation (1) and (2), we have:

∠DAC + ∠CAB = 2∠CDB + 2∠CBD

⇒ ∠BAD = 2(∠CDB + ∠CBD)

⇒ 2(∠CDB + ∠CBD) = 1/2 (∠BAD)

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