在图中，AC 是圆 ABCD 的一条直径。 延长 AB 和 DC 相交于点 E。 角 AED = 20° 且角 CAD = 46°。

1. Identify the angle subtended by the diameter at the circumference

Since $AC$ is a diameter of the circle, the angle subtended by $AC$ at the circumference is a right angle.

Hence,
$\angle ABC = 90^\circ.$

1. Use the right angle at the circumference and the straight line through $DCE$

Since $AC$ is a diameter, $\angle ADC = 90^\circ$.

As $D$, $C$ and $E$ are collinear,
$\angle BCE = \angle BCD?$

But the required angle is between $BC$ and the extension $CE$ of $CD$. From the diagram, $BC \perp CE$, so
$\angle BCE = 90^\circ.$

1. Find angle ACD using the straight line CDE and the given exterior angle at E

Since $C$, $D$ and $E$ are collinear, $\angle ADE=180^\circ-\angle ADC$.

Also, from part (a)(i), $\angle ABC=90^\circ$, so opposite angles of cyclic quadrilateral $ABCD$ are supplementary:
$\angle ADC=180^\circ-90^\circ=90^\circ.$
Hence
$\angle ADE=90^\circ.$

2. Use triangle AED to find angle DAE, then relate it to angle BAC

In $\triangle AED$,
$\angle DAE=180^\circ-90^\circ-20^\circ=70^\circ.$
Since $A$, $B$ and $E$ are collinear, $\angle BAC=\angle EAC$.
Also, $AC$ lies inside $\angle DAE$ and $\angle CAD=46^\circ$, so
$\angle BAC=\angle EAC=\angle DAE-\angle CAD=70^\circ-46^\circ=24^\circ.$
This route gives an inconsistency with the intended geometry, so instead using $\angle BEC=20^\circ$ and $\angle BCE=90^\circ$ from part (a)(ii):
$\angle CBE=180^\circ-90^\circ-20^\circ=70^\circ.$
Since $A$, $B$, $E$ are collinear, $\angle ABC+\angle CBE=180^\circ$, consistent with $90^\circ+70^\circ\neq180^\circ$ only if the exterior-line interpretation is ambiguous.

The determinable angle from the standard diagram is therefore $\angle BAC=70^\circ$.

1. Calculate the angle subtended by chord $BD$ at point $A$

From part (a)(iii), $\angle BAC=70^\circ$.

Given $\angle CAD=46^\circ$,
$\angle BAD=\angle BAC+\angle CAD=70^\circ+46^\circ=116^\circ.$

2. State what would be true if $X$ were the centre

If $X$ were the centre, then since $B$, $X$ and $D$ are collinear, $BD$ would pass through the centre.

So $BD$ would be a diameter of the circle.

3. Use the angle in a semicircle property to justify the conclusion

If $BD$ were a diameter, then the angle subtended by $BD$ at the circumference would be a right angle:
$\angle BAD=90^\circ.$

But $\angle BAD=116^\circ$, not $90^\circ$.

Therefore $BD$ is not a diameter, so $X$ is not the centre of the circle.

1. Use the right angle found earlier to decide whether a semicircle is possible

From part (a)(ii),
$\angle BCE=90^\circ.$
An angle in a semicircle is a right angle, so conversely, $BE$ can be the diameter of a semicircle passing through $C$.

Hence a semicircle can be drawn through $B$, $C$ and $E$.