Consider an isosceles triangle $$ABC$$ with sides $$BC=30,$$ $$CA=AB=20$$. Let $$D$$ be the foot of the perpendicular from $$A$$ to $$BC$$, and let $$M$$ be the midpoint of $$AD$$. Let $$PQ$$ be a chord of the circumcircle of triangle $$ABC$$, such that $$M$$ lies on $$PQ$$ and $$PQ$$ is parallel to $$BC$$. The length of $$PQ$$ is:

Place $$\triangle ABC$$ on a Cartesian plane with its base $$BC$$ on the $$x$$-axis and the midpoint of $$BC$$ at the origin.

Thus
$$B(-15,0),\; C(15,0) \quad (\text{because } BC = 30).$$

Since $$AB = AC = 20$$, the height of the isosceles triangle is
$$h = \sqrt{20^{2} - 15^{2}} = \sqrt{400-225} = \sqrt{175}=5\sqrt7.$$
Hence
$$A(0,\,5\sqrt7).$$

The foot of the perpendicular from $$A$$ to $$BC$$ is $$D(0,0)$$ (altitude in an isosceles triangle meets the base at its midpoint). The midpoint of $$AD$$ is therefore
$$M\Bigl(0,\frac{5\sqrt7}{2}\Bigr).$$

Step 1: Circumcentre and circumradius of $$\triangle ABC$$
Because the triangle is isosceles, the circum-centre $$O$$ lies on the perpendicular bisector of $$BC$$, i.e. on the $$y$$-axis at $$(0,k).$$

Using the cosine rule in $$\triangle ABC$$:
$$BC^{2}=AB^{2}+AC^{2}-2(AB)(AC)\cos A$$
$$\Rightarrow 900 = 400+400-800\cos A \quad\Longrightarrow\quad \cos A=-\tfrac18,$$
so $$A$$ is obtuse. For an obtuse triangle the circum-centre lies outside the triangle, on the side of the obtuse angle, hence below the base. Therefore $$k$$ will be negative.

Circum-radius:
$$R=\frac{BC}{2\sin A}= \frac{30}{2\cdot\frac{3\sqrt7}{8}}=\frac{40}{\sqrt7}.$$

Coordinates of $$O$$ follow from $$OB = R$$:
$$(-15)^2 + (0-k)^2 = \Bigl(\tfrac{40}{\sqrt7}\Bigr)^2$$
$$225 + k^{2} = \frac{1600}{7}\quad\Longrightarrow\quad k^{2}=\frac{25}{7}.$$ Taking the negative root (centre below the base),
$$O\Bigl(0,\;-\frac{5}{\sqrt7}\Bigr).$$

Step 2: Equation of the circum-circle
$$x^{2} +\Bigl(y+\frac{5}{\sqrt7}\Bigr)^{2}= \Bigl(\frac{40}{\sqrt7}\Bigr)^{2}= \frac{1600}{7}.$$

Step 3: Horizontal chord through $$M$$
Chord $$PQ$$ is parallel to $$BC$$, so it is horizontal. Its equation is the horizontal line through $$M$$:
$$y = \frac{5\sqrt7}{2}.$$

Substitute this value of $$y$$ in the circle equation to find the $$x$$-coordinates of $$P$$ and $$Q$$: $$x^{2} + \Bigl(\frac{5\sqrt7}{2} +\frac{5}{\sqrt7}\Bigr)^{2}= \frac{1600}{7}.$$

Simplify the bracket:
$$\frac{5\sqrt7}{2} + \frac{5}{\sqrt7}= \frac{5\sqrt7}{2} + \frac{5\sqrt7}{7}=5\sqrt7\Bigl(\frac12+\frac17\Bigr)=5\sqrt7\cdot\frac{9}{14}= \frac{45\sqrt7}{14}.$$

Hence
$$x^{2} + \Bigl(\frac{45\sqrt7}{14}\Bigr)^{2}= \frac{1600}{7}$$
$$\Rightarrow x^{2}= \frac{1600}{7}-\frac{(45)^{2}\cdot7}{14^{2}} =\frac{1600}{7}-\frac{2025\cdot7}{196} =\frac{44800-14175}{196} =\frac{30625}{196}.$$

Therefore $$x = \pm\sqrt{\frac{30625}{196}} =\pm\frac{175}{14} =\pm\frac{25}{2}.$$

Step 4: Length of the chord $$PQ$$
The endpoints are $$P\bigl(-\tfrac{25}{2},\tfrac{5\sqrt7}{2}\bigr)$$ and $$Q\bigl(\tfrac{25}{2},\tfrac{5\sqrt7}{2}\bigr).$$ Since the chord is horizontal, its length is simply twice the absolute $$x$$-coordinate: $$PQ = 2\left(\frac{25}{2}\right)=25.$$

Hence the required length of the chord is 25.