In the given circle with diameter AD, and point B on the circumference, where BC is perpendicular to AD. If ABC = 41^, find D.

Given:

• AD is the diameter of the circle with center O.
• B is a point on the circumference.
• BC $\perp$ AD, which means $\angle BCA = 90^\circ$ and $\angle BCD = 90^\circ$.
• $\angle B_1 = 41^\circ$. This refers to $\angle ABC$.

Since AD is the diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$. Therefore, $\angle ABD = 90^\circ$.
In $\triangle BCD$:
We know $\angle BCD = 90^\circ$ (given BC