A、B、C、D 是圆上的点。 直线 PCQ 是圆的切线。
直线 QE 平分角 AEC 且平行于直线 CD。

1. Identify the two line segments as radii of the circle.

$E$ is the centre of the circle, and $C$ and $D$ lie on the circumference.

Therefore, $EC$ and $ED$ are radii of the circle.

2. Use the property that radii of the same circle are equal.

All radii of the same circle are equal.

Hence, $EC = ED$.

1. Find the radius of the circle.

Since $AD$ is the diameter,
$r=\frac{28}{2}=14\text{ cm}.$

2. Use the tangent-radius property to find $\angle FCE$.

Since $QC$ is a tangent at $C$, $EC \perp QC$.

So
$\angle QCE=90^\circ.$
Given $\angle QCF=61^\circ$,
$\angle FCE=90^\circ-61^\circ=29^\circ.$

3. Find the central angle $\angle CED$ subtending chord $CD$.

In $\triangle ECF$, $EC=EF$ since both are radii.

Hence
$\angle EFC=\angle FCE=29^\circ.$
So
$\angle CEF=180^\circ-29^\circ-29^\circ=122^\circ.$
Since $EF \parallel CD$,
$\angle ECD=180^\circ-122^\circ=58^\circ.$
From part (a), $EC=ED$, so $\triangle ECD$ is isosceles.

Thus
$\angle EDC=58^\circ.$
Therefore
$\angle CED=180^\circ-58^\circ-58^\circ=64^\circ.$

4. Calculate the area of the minor segment $CD$.

Area of segment $CD$
$=\text{area of sector }CED-\text{area of }\triangle CED.$

$=\frac{64}{360}\pi(14)^2-\frac12(14)(14)\sin64^\circ$

$=21.386\ldots$

$\approx 21.4\text{ cm}^2.$