图中，$AC$ 是与圆相交于点 $A$ 和 $D$ 的一条直线。点 $B$ 在圆上，且 $BC$ 是圆的切线。点 $E$ 在 $AC$ 上，使得直线 $BE$ 平分角 $ABD$。

1. State the geometric reason for the equality of the angles

The angle between the tangent $BC$ and the chord $BD$ is equal to the angle in the alternate segment, which is angle $BAD$. This is known as the alternate segment theorem.

1. Express angle EBC in terms of its component angles

$\angle EBC = \angle EBD + \angle DBC$

2. Use the given angle bisector property and the result from part (i)

Since $BE$ bisects $\angle ABD$, $\angle EBD = \angle ABE$.
From (i), $\angle DBC = \angle BAD$ (alternate segment theorem).
Therefore, $\angle EBC = \angle ABE + \angle BAD$.

3. Relate the sum of angles to angle BEC using the exterior angle of a triangle

In $\triangle ABE$, the exterior angle at $E$ is $\angle BEC$.
$\angle BEC = \angle ABE + \angle BAE$ (exterior angle of a triangle).
Since $\angle BAE$ is the same as $\angle BAD$, we have $\angle BEC = \angle ABE + \angle BAD$.
Thus, $\angle EBC = \angle BEC$.

1. Identify the isosceles triangle and state the reason based on the result from part (ii)

From (ii), we have $\angle EBC = \angle BEC$.
Since two angles in triangle $EBC$ are equal, the sides opposite to them are equal ($BC = EC$).
Therefore, triangle $EBC$ is an isosceles triangle.