Diagonals in an inscribed quadrilateral can simultaneously be angle bisectors of triangles involving their midpoints

Question

This is follow-up of an interesting question asked some days ago on this site, that the asker has erased 24 hours later.

The initial question was (as reflected in my title) in terms of an inscribed quadrilateral $ABCD$. Its formulation had the drawback of being "not constructive" (it was for example difficult to make an exact figure) ; here is the same problem presented under a different form.

The figure above features an acute triangle $ACD$, $M$ the midpoint of $AC$, $E$ the symmetrical point of $D$ with respect to line bisector of $AC$, $B$ the intersection point of $EM$ with the circumscribed circle to $ACD$ (in this way, constraint $\angle DMA = \angle AMB$ is fulfilled). Let $N$ be the midpoint of $BD$. The objective is to show that $NB$ is the angle bisector of triangle $ANC$.

My question :

I have an analytic geometry solution (see below).

Is there a "synthetic geometry" solution ?

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Analytic geometry solution :

Take coordinate axes as shown on the figure with $A=(-1,0)$ and $B=(1,0)$. Let $D=(x_0,y_0)$ ; therefore $E(-x_0,y_0)$.

As $\vec{MB}=-k \vec{ME}$, with a certain value of $k$, the coordinates of $B$ are $(kx_0,-ky_0)$.

Computing the power of $M$ with respect to the circle in 2 different ways : $$\underbrace{\overline{MB}.\overline{ME}}_{-k ME^2}=\underbrace{\overline{MA}.\overline{MC}}_{-1}$$

As a conclusion :

$$k=\frac{1}{x_0^2+y_0^2}.\tag{1}$$

Besides, it isn't difficult to show that the intersection point $P$ of $BD$ with $AC$ can be expressed like this :

$$P=\left(\frac{2k}{1+k}x_0,0\right)$$

Let us apply the angle bissector theorem in the reverse way (criteria for a cevian to be an angle bissector) :

$$\frac{NA}{NC}\stackrel{?}{=}\frac{PA}{PC}$$

giving :

$$\frac{((1+k)x_0+2)^2+(1-k)^2y_0^2}{((1+k)x_0-2)^2+(1-k)^2y_0^2}\stackrel{?}{=}\frac{(2kx_0+1+k)^2}{(2kx_0-(1+k))^2},$$

which is easily verified by a CAS with the expression of $k$ given in (1).

(End of proof)

Remark : (following the answer of @sirous) : Let $F$ be the intersection of diagonal $AD$ with the line bissector of $AC$ (vertical axis). It suffices to show that $F$ belongs to the circumcircle of $ACN$.

Answer 1

Let $\angle CDE=\alpha, ~\angle CDM=\beta$ and $\angle MDB=\gamma$.

Simple angle chasing (using $MD=ME$ and $AC||DE$) leads to $\angle ADB=\beta$.

Hence, $DB$ is the $D$-symmedian of $\triangle ADC$. Using the powerful Lemma 4.26 (f) in EGMO, $AC$ is the $A$-symmedian of $\triangle ABD$ and also the $C$-symmedian of $\triangle CBD$.

Finally, angle chasing leads to $\angle ANB=\angle CNB=2\beta+\gamma$.

Answer 2

May be this idea works: As can be seen in figure the circle passing the center of circumcircle of quadrilateral ABCD and the ends of it's diagonals passes through mid points of non corresponding diagonals, because diagonals are the common chord of the each two circles. So OF and OE are diameters of these circles which meet the circles at points H and G where the extensions of CA and DB meet the circle. This indicates that FG and EH must be the bisectors of angle AFC and BED respectively.