Germany TST

Solution:

If we set $\text{= DAB}$, then we must also have $\text{ABD =}$, since triangle $ABD$ was assumed to be isosceles. From this it follows that $\text{BDA = 180 - 2}$, and by means of the inscribed angle theorem it follows that $\text{BCE = BCA = 180 - 2}$.

Next we want to express the angle $BIE$ in terms of $\varphi$: Since the lines $BI$ and $EI$ are angle bisectors in triangle $BCE$, we have
$$\text{BIE = 180 - ( IEB + EBI) = 90 + 1 2 (180 - CEB + EBC ) = 90 + 1 2 BCE,}$$
which, combined with the previous result, gives
$$\text{BIE = 90 + 1 2 (180 - 2 ) = 180 - .}$$

Since $BIEN$ is a convex cyclic quadrilateral, this implies $\text{CNB =}$. Since by the inscribed angle theorem over the chord $\overline{AD}$ we also have $\text{ACD =}$, it follows that $CD \parallel BN$ must hold.

Now extend the segment $\overline{BN}$ beyond $N$ until it meets the circumcircle of the quadrilateral $ABCD$ again at $Q$. By applying the inscribed angle theorem over the chord $\overline{BD}$, we obtain $\text{DQB =}$, which in turn implies $CN \parallel QD$.

Since both pairs of opposite sides of the quadrilateral $CNQD$ are thus parallel, it must be a parallelogram. Hence $|CD| = |NQ|$, and as soon as we substitute this into the equation
$$|AN| \cdot |NC| = |BN| \cdot |NQ|$$
which follows from the power of a point (intersecting chords theorem), the claim is proven.