IRL ABooklet

Let $D$ be the intersection point of $BF$ and $CE$. The circumcircle $\Omega$ of triangle $ABC$ has centre $O$ and radius $r = |ON|$. Let $G$ be the second intersection point of the circumcircle of triangle $ABF$ and the line $EF$. We then have $\angle CAB = \angle BGE$ and
$$|EB| \cdot |EA| = |EG| \cdot |EF| \qquad (12)$$
Because $ABDC$ is cyclic, we have $\angle BDE = \angle CAB$ and so $\angle BDE = \angle BGE$ which implies that $BEGD$ is cyclic. This gives
$$|FD| \cdot |FB| = |GF| \cdot |EF|. \qquad (13)$$
Adding (12) and (13) gives
$$|EB| \cdot |EA| + |FD| \cdot |FB| = |EF|^2. \qquad (14)$$
Considering the power of the points $E$ and $F$ with respect to the circle $\Omega$, we obtain
$$|EB| \cdot |EA| = |OE|^2 - r^2 \quad \text{and} \quad |FD| \cdot |FB| = |OF|^2 - r^2.$$
Adding these together yields
$$|EB| \cdot |EA| + |FD| \cdot |FB| = |OE|^2 + |OF|^2 - 2r^2. \qquad (15)$$
To prove that $\triangle MNO$ is right angled, we now need a formula for the median $OM$ of triangle $EFO$. Such a formula is well known and can be obtained from the Cosine Rule as follows. Let $\theta = \angle OME$, then $\angle FMO = 180^{\circ} - \theta$ and $\cos(180^{\circ} - \theta) = -\cos(\theta)$. The Cosine Rule for triangles $FMO$ and $OME$ gives
$$\begin{aligned} |OF|^2 &= |FM|^2 + |OM|^2 + 2|FM| \cdot |OM| \cos(\theta) \\ |OE|^2 &= |EM|^2 + |OM|^2 - 2|EM| \cdot |OM| \cos(\theta) \end{aligned}$$
Adding these together and taking into account that $|EM| = |FM|$, we obtain
$$|OF|^2 + |OE|^2 = 2|OM|^2 + 2|EM|^2. \qquad (16)$$
Because $|EF| = 2|EM|$, (14), (15) and (16) give us
$$4|EM|^2 = |EF|^2 = |OE|^2 + |OF|^2 - 2r^2 = 2|OM|^2 + 2|EM|^2 - 2r^2,$$
and we obtain $2|EM|^2 = 2|OM|^2 - 2r^2$. Using $r = |ON|$ and $|EM| = |MN|$, this can be rewritten as
$$|MN|^2 + |ON|^2 = |OM|^2$$

and this means that triangle $MNO$ has a right angle at $N$.