In the diagram above; O is the centre of the circle and |BD| = |DC|. If ?DCB = 35°, find ?BAO.

We can start by using the fact that the angle at the center of the circle is twice the angle at the circumference that subtends the same arc. Therefore, since |BD| = |DC|, then angle CBD = angle BDC. Also, angle COB is equal to 2*angle BDC because it subtends the same arc as angle BDC. Therefore, angle COB = 2*angle CBD. Since triangle AOB is isosceles (|OA| = |OB|), then angle BAO = angle BOA. Using the fact that the angles in a triangle sum to 180 degrees, we can write: angle BAO + angle BOA + angle COB = 180 Substituting in the expressions we derived earlier, we get: angle BAO + angle BAO + 2*angle CBD = 180 Simplifying, we get: 2*angle BAO + 2*angle CBD = 180 angle BAO + angle CBD = 90 Now we can use the fact that angle DCB = 35 degrees and |BD| = |DC| to find angle CBD: angle DCB + angle BDC + angle CBD = 180 35 + 35 + angle CBD = 180 angle CBD = 110 Substituting this value back into the earlier equation, we get: angle BAO + 110 = 90 angle BAO = 90 - 110 = -20 Since angle BAO cannot be negative, we made an error along the way. Double-checking our work, we realize that we made a mistake when substituting in the value for angle COB. It should be equal to 2*angle DCB, not 2*angle BDC. Therefore: angle COB = 2*angle DCB = 2*35 = 70 Substituting this value back into the earlier equation, we get: angle BAO + 70 = 90 angle BAO = 20 Therefore, the answer is 20 degrees.