The three sides of the trapezoidal section of most economical section are tangential to the semi-circle described on the water line. This is proved as

Let Fig.shows the trapezoidal channel of most economical section

Let $\theta=$ angle made by the the sloping sides with horizontal

Draw $O F$ perpendicular to the sloping side $A B$.

$\triangle O A F$ is a right-angled triangle and $\angle O A F=\theta$

$\begin{aligned} \sin \theta &=\frac{OF}{OA} \\ OF &= OA\sin \theta......(i) \end {aligned}$

In $\triangle A E B, $

$\begin {aligned} \sin \theta &=\frac{A E}{A B} \\ &=\frac{d}{\sqrt{d^{2}+n^{2} d^{2}}}\\ &=\frac{d}{d \sqrt{1+n^{2}}}\\ &=\frac{1}{\sqrt{1+n^{2}}} \end{aligned}$

Substituting $\sin \theta=\frac{1}{\sqrt{1+n^{2}}}$ in equation (i), we get

$ \begin{aligned} O F &=A O \times \frac{1}{\sqrt{1+n^{2}}}....(ii) \\ A O &=\text { half of top width } \\ &=\frac{b+2 n d}{2}\\ &=d \sqrt{n^{2}+1} \end{aligned} $

But Substituting this value of $A O$ in equation (ii)

$\begin{aligned} O F &=\frac{d \sqrt{n^{2}+1}}{\sqrt{n^{2}+1}}\\ &=d \text { depth of flow } \end{aligned}$

Thus, if a semi-circle is drawn with $O$ as centre and radius equal to the depth of flow $d$, the three sides of most economical trapezoidal section will be tangential to the semi-circle.