This question has already been asked yesterday by the user @anonymus. I tried to solve it unsuccessfully after leaving a longer comment to persuade the $OP$ to include personal thoughts in the post. Since nothing has happened up to this moment, I voted to close it and ask the same question here including my attempt.
Here it goes:
Let $ABCD$ be a quadrilateral inscribed in a circle, where $|DC|<|AB|$ and $DC\nparallel AB$. Let $X$ be the intersection point of the diagonals $\overline{AC}$ and $\overline{BD}$. And $Y$ be the foot of the perpendicular from $X$ on the edge $\overline{AB}$. If $XY$ bisects the angle $\measuredangle{DYC}$, prove that $\overline{AB}$ is the diameter of the (circum)circle, i.e. $ABCD$ is a semicyclic quadrilateral.
My attempt:
If $XY$ bisects $\measuredangle DYC$, then $\measuredangle DYX=\measuredangle XYC$.
$$\color{red}{\measuredangle AYD}=90^{\circ}-\measuredangle DYX=90^{\circ}-\measuredangle XYC=\color{red}{\measuredangle CYB}$$
$$\measuredangle C'YA=\measuredangle AYD$$
When drawing, I noticed that $X$ is the center of the circle inscribed in $\Delta DYC$
$$\implies\color{green}{\measuredangle CDB=\measuredangle BDY}\;\&\;\color{blue}{\measuredangle YCA=\measuredangle ACD}$$
I tried using the following:
$$\color{purple}{\Delta ABX\sim\Delta CDX}\;\&\;\Delta AXD\sim\Delta CXB$$
My reasoning is circular.
I'm not sure if I should already assume $\color{brown}{\measuredangle{BDA}=\measuredangle{BCA}=90^{\circ}}$.
Then there's no point in stating that $BCXY$ is also a cyclic quadrilateral.
How can I continue and improve what I've written so far? Thank you in advance!
Update:
For all those wondering, thanks to @Blue in the comment section, I will read more on the topic: Incircle and excircles of a triangle.
Picture:
Best Answer
Here's an approach that may be unnecessarily-complicated.
In the figure, $\angle BAC\cong\angle BDC$ and $\angle ABD\cong\angle ACD$, since each pairs of angles subtend the same arcs. A little angle-chasing gives $\angle YCA=90^\circ-\alpha-\theta$ and $\angle YDB=90^\circ-\beta-\theta$.
By the trigonometric form of Ceva's Theorem (see an alternative below), we have $$1 = \frac{\sin\angle CYX}{\sin\angle XYD}\cdot\frac{\sin\angle DCX}{\sin\angle XCY}\cdot\frac{\sin\angle YDX}{\sin\angle XDC} = 1\cdot\frac{\sin\beta}{\sin(90^\circ-\alpha-\theta)}\cdot\frac{\sin(90^\circ-\beta-\theta)}{\sin\alpha} \tag{1}$$ so that $$\sin\alpha\cos(\alpha+\theta) = \sin\beta\cos(\beta+\theta) \quad\to\quad \sin(\alpha-\beta)\cos(\alpha+\beta+\theta) = 0 \tag{2}$$ Since $\alpha$, $\beta$, $\theta$ are positive and acute, we have either that $\alpha=\beta$ or $\alpha+\beta+\theta=90^\circ$. The former would make $\overline{AB}\parallel\overline{CD}$, which violates an assumption; thus, the latter holds. Substituting into the expressions for $\angle YDB$, we find this equal to $\alpha$, and is thus also equal to $\angle YDX$. This makes $\square XYAD$ a cyclic quadrilateral whose opposite angles at $Y$ and $D$ must be supplementary. The result follows. $\square$
Note. I like to throw in trigonometric Ceva whenever possible, because I don't think it gets enough attention, but it isn't necessary to get to $(2)$.
Instead, we can define, say, $x := |XY|$ and use straightforward trig to give expressions in $x$, $\alpha$, $\beta$, $\theta$ for the lengths of the subsegments of the diagonals of $\square ABCD$. Then, we can use the similarity $\triangle AXB\sim\triangle DXB$ (or, equivalently, the chord-chord aspect of the Power of a Point theorem) to write $$|XA|\cdot|XC| = |XB|\cdot|XD| \tag{3}$$ and manipulate the result into $(2)$.