in the figure, chord AB of circle with centre O, is produced to C such that BC = OB . CO is joined and produced to meet the circle at D.If angle ACD = y and angle AOD = x,show that x = 3y.
Answers
sorry for the drawing
Answer:
The relation between ∠AOD and ∠ACD is x = 3y
Step-by-step explanation:
According the given question, the figure can be drawn as(refer to the attachment)
given,
AB is the chord of the circle (center O)
BC=OB
Since the sides BC and OB is equal, we know that, angles opposite to equal sides are equal.
hence ∠OCB=∠BOC=y°
From the figure, ∠OBA is the exterior angle of the triangle ΔBCO.
With the exterior angle property of the triangle i.e. sum of opposite interior angels is equal to exterior angle
∴ ∠OBA =∠BOC+∠OCB
= y°+y° = 2y°
∠OBA=2y°
OA=OB (the radii of the circle)
Hence, ∠OAB=∠OBA (angles opposite to equal sides are equal. ).
⇒∠OAB= 2y°
we need to find, ∠AOD, which is exterior angle of ΔAOC
∴ ∠AOD=∠OAB+∠OCA
x°= 2y°+y°
∴ x°= 3y°
Hence Proved
