geometry
Let $ABC$ be an acute triangle with altitudes $BB_1$ and $CC_1$. Let $B_2$ and $C_2$ be the midpoints of $AC$ and $AB$, respectively. Find all possible values of $\angle BAC$ if $B_1C_2$ and $B_2C_1$ are perpendicular.
I have angle chased this problem many times, but I always keep getting $\angle BAC = 90^{\circ}$, contrary to the hypotesis of the problem that $ABC$ is an acute triangle.
As in the above drawing let $\angle DAL=\angle OAC=a$
And let be the intersection of the angle bisector, perpendicular bisector and the altitude be $O$
Let the perpendicular bisector be $DK$
$\angle DAO=\angle OAH=a and \angle AOH=\angle DOA=90-a$
Then $\triangle DAO \equiv \triangle OAH (A.A.S)$
With this you can get that $AD=AH$ and $OD=OH$
Since $DK$ is the perpendicular bisector of $AB$ $AD=DB$1
Now draw $RH$ and also the median respective to the right angle in a right angles triangle is equal to the half of the hypotenuse
$AD=AH$
And that brings us to $AD=AH=DH$
Since $\triangle ADH$ is equilateral
$\angle BAC=\angle ADH=\angle DHA=60$
Best Answer
Add lines $BB_3$ and $CC_3$ such that $BB_3 // C_2B_1$ and $CC_3 // B_2C_1$.
By intercept theorem, $AB_1 = B_1B_3$ (and also $AC_1 = C_1C_3$)
Together with $BB_1$ is an altitude, we have $BA = BB_3$.
Similarly, $CA = CC_3$
Thus, the red marked angles are equal and $= 180^0 – 2\angle A$
By angle sum of triangle, $180^0 = \angle BKC + x + y$
$180^0 = \angle BKC + [\angle B – (180^0 – 2A)] + [\angle C – (180^0 – 2A)]$
:
$360^0 = \angle BKC + 3\angle A$
By corresponding angles between parallels, $\angle BKC = \angle B_1HC1 = \angle B_3KC_3 = 90^0$
∴ $\angle A$ has to be $90^0$
This means that requiring $\triangle ABC$ to be acute should be re-phrased as $\triangle ABC$ is non-obtuse.