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 $$|EB|\cdot |EA|=|EG|\cdot |EF|(12)$$ Because $ABDC$ is cyclic, we have $\angle BDE=\angle CAB$ and so $\angle BDE=\angle BGE$ which implies that $BEGD$ is cyclic. This gives $$|FD|\cdot |FB|=|GF|\cdot |EF|.(13)$$ Adding (12) and (13) gives $$|EB|\cdot |EA|+|FD|\cdot |FB|=|EF{|}^2.(14)$$ Considering the power of the points $E$ and $F$ with respect to the circle $\Omega$, we obtain $$|EB|\cdot |EA|=|OE{|}^2-r^2\text{and}|FD|\cdot |FB|=|OF{|}^2-r^2.$$ Adding these together yields $$|EB|\cdot |EA|+|FD|\cdot |FB|=|OE{|}^2+|OF{|}^2-2r^2.(15)$$ To prove that $△MNO$ 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 $$\begin{array}{rl}|OF{|}^2& =|FM{|}^2+|OM{|}^2+2|FM|\cdot |OM|\cos (\theta )\\ |OE{|}^2& =|EM{|}^2+|OM{|}^2-2|EM|\cdot |OM|\cos (\theta )\end{array}$$ Adding these together and taking into account that $|EM|=|FM|$, we obtain $$|OF{|}^2+|OE{|}^2=2|OM{|}^2+2|EM{|}^2.(16)$$ 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 $N$.