3. Let $ABCD$ be a rhombus with the shorter diagonal $BD$ and $E$ an internal point on its side $CD$, which lies on the circumcircle of triangle $ABD$. Determine the size of its interior angle at vertex $A$, given that the circumcircles of triangles $ACD$ and $BCE$ have exactly one common point.
Let's denote the sought internal angle at vertex $A$ of the considered rhombus as $\alpha$. Further, let $k$ be the circumcircle of triangle $A C D$ and $l$ the circumcircle of triangle $B C E$.

From the conditions of the problem, it follows that $A B E D$ is an isosceles trapezoid with the shorter base $E D$. Therefore, $|E B|=|D A|=|C B|$, which means that triangle $B C E$ is isosceles with base $C E$.

According to the problem statement, point $C$ is the only common point of circles $k$ and $l$, so there exists a common tangent $t$ to both circles at this point (Fig. 1). Since $E$ is an internal point of side $C D$ (a chord of circle $k$), both circles have an internal tangency at point $C$. Consequently, the tangent $t$ forms the same segment angle with the chord $C D$ of circle $k$ and the chord $C E$ of circle $l$, as indicated in the figure. This means that the corresponding inscribed angles of the mentioned chords of circles $k$ and $l$ are equal, i.e., $|\angle C A D|=|\angle C B E|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c37ab977ba4d4a3e6d1eg-3.jpg?height=648&width=920&top_left_y=1104&top_left_x=571)

Fig. 1

Since the diagonal $A C$ is the bisector of the internal angle at vertex $A$ in the rhombus $A B C D$, we have $|\angle C A D|=\frac{1}{2} \alpha$, while the isosceles triangle $C E B$ has base angles of $|\angle B E C|=|\angle B C E|=|\angle B A D|=\alpha$, so $|\angle C B E|=180^{\circ}-2 \alpha$. Together, we get the equation for the angle $\alpha$:

$$
180^{\circ}-2 \alpha=\frac{1}{2} \alpha
$$

The solution to this equation is $\alpha=72^{\circ}$.

GUIDING PROBLEMS:

N1. A circle is circumscribed around triangle $A B C$ with internal angles $\alpha, \beta, \gamma$. Tangents are drawn to this circle at points $A, B, C$. Determine the sizes of the internal angles of the triangle bounded by these tangents. Analyze different shapes of triangle $A B C$. [For example, for an acute triangle with differently sized internal angles, the angles of the new triangle are $180^{\circ}-2 \alpha, 180^{\circ}-2 \beta, 180^{\circ}-2 \gamma$.]

N2. Prove that a circle can be circumscribed around an isosceles trapezoid and that a circle cannot be circumscribed around any other trapezoid.

N3. In an isosceles trapezoid $A B C D$, $|B C|=|C D|=|D A|$ and $|\angle D A B| = |\angle A B C| = 36^{\circ}$. On the base $A B$, a point $K$ is given such that $|A K|=|A D|$. Prove that the circles circumscribed around triangles $A K D$ and $K B C$ have an external tangency. [53-B-I-2]

N4. A trapezoid $A B C D$ with acute angles at the base $A B$ is given. On this base, there exists a point $E$ such that the circles circumscribed around triangles $A E D$ and $E B C$ have an external tangency. Prove that point $E$ lies on the circle circumscribed around triangle $C D V$, where $V$ is the intersection of lines $A D$ and $B C$. [53-B-S-3]

N5. A parallelogram $A B C D$ with an obtuse angle $A B C$ is given. On its diagonal $A C$ in the half-plane $B D C$, choose a point $P$ such that $|\angle B P D|=|\angle A B C|$. Prove that the line $C D$ is tangent to the circle circumscribed around triangle $B C P$ if and only if segments $A B$ and $B D$ are congruent. [59-A-II-2]