The following is Lemma $1.44$ in Euclidean Geometry in Mathematical Olympiads by Evan Chen.
Let $ABC$ be an acute triangle. Let $BE$ and $CF$ be altitudes of $\triangle ABC$, and denote by $M$ the midpoint of $BC$. Prove that $ME, MF,$ and the line through $A$ parallel to $BC$ are all tangents to $(AEF)$.
$\textbf{MY ATTEMPT -}$
Let $H$ be the orthocentre of $\triangle ABC.$ It is well known that $AFHE$ is cyclic. Let $O$ be the centre of its circumcircle$(AFHE)$.
Since $\angle BFC = \angle BEC = 90^{\circ}$, $BFEC$ is cyclic and $(BFEC)$ has centre $M$. Thus $MB=MF=ME=MC.$
Let $\angle BAC = \alpha, \angle ABC = \beta, \angle BCA = \gamma$.
$$\angle FCB = \angle FEB = 90^{\circ}-\beta$$
Since $MB = ME$,
$$\angle MBE = \angle BEM = 90^{\circ}-\gamma$$
By Inscribed Angle Theorem,
$$\angle FOE = 2\angle FAE = 2\alpha \implies \angle OEF = 90^{\circ}-\alpha$$ Thus, $$\angle OEM = \angle OEF + \angle FEB + \angle BEM = 90^{\circ}$$
As $AFHE$ is cyclic,
$$\angle FAH = \angle FEH = \angle FEB = \angle FCB = 90^{\circ}-\beta$$
As $BC \parallel AD$,
$$\angle DAB = \angle CBA = \beta$$
Thus, $$\angle DAO = 90^{\circ}$$ $\blacksquare$
$\textbf{QUESTION -}$
Although the lemma can be proved using ordinary angles, the hint supplied with the question is,
$$\measuredangle FEM = \measuredangle FEB + \measuredangle BEM = \measuredangle FEB + \dots?$$
This indicates that the problem can also be done using directed angles, but I am finding it difficult to convert the stated proof into directed angles since it requires halving a directed angle after applying the Inscribed Angle Theorem, which is wrong.
Can anyone explain how the question can be solved via the supplied hint?
Any help on the topic would be appreciated
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