Let D be the foot of the A-altitude, AD = h. WLOG, assume DB < DC. By similar right angle triangles or by the power of the point D to the circumcircle of the right angle triangle $\triangle ABC$, $DB \cdot (a - DB) = h^2$. Let $PQ = \frac a n$ be the defined segment containing the hypotenuse midpoint, such that the points B, D, P, Q, C follow on the hypotenuse BC in this order. Denote $\phi = \angle PAD,\ \psi = \angle QAD$. $\tan \alpha = \tan(\psi - \phi) = \frac{\tan \psi - \tan \phi}{1 + \tan \psi \cdot \tan \phi}$ $\tan \psi = \frac{QD}{h} = \frac{QB - DB}{h},\ \ \tan \psi = \frac{PD}{h}= \frac{PB - DB}{h}$ $\tan \psi - \tan \phi = \frac{QD - PD}{h} = \frac{PQ}{h} = \frac{a}{nh}$ $1 + \tan \psi \cdot \tan \phi = 1 + \frac{(QB - DB)(PB - DB)}{h^2} =$ $= 1 + \frac{QB \cdot PB - (QB + PB) DB + DB^2}{h^2} = 1 + \frac{a\frac{n + 1}{2n} \cdot a\frac{n - 1}{2n} - a DB + DB^2}{h^2} =$ $= 1 + \frac{a^2\frac{n^2 - 1}{4n^2} - DB (a - DB)}{h^2} = \frac{a^2}{h^2} \cdot \frac{n^2 - 1}{4n^2}$ As a result, $\tan \alpha = \frac{a}{nh} \cdot \frac{4n^2h^2}{a^2(n^2 - 1)} = \frac{4nh}{a(n^2 - 1)}$

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We can let each of the n segments have length 1, then a=n. Let the middle segment be XY. The area of $ \triangle AXY$ is h/2. So $ h = AX \cdot AY \cdot \sin{\alpha}$ By the Law of cosines on $ \triangle AXY$, $ AX^2 + AY^2 - 1 = 2AX \cdot AY \cdot\cos{\alpha}$ Dividing, we have $ \tan{\alpha} = \frac {2h}{AX^2 + AY^2 - 1}$ So it remains to prove $ \frac {2h}{AX^2 + AY^2 - 1} = \frac {4h}{n^2 - 1}$ (since a=n) $ AX^2 + AY^2 - 1 = (n^2 - 1)/2$ We now use Stewart's twice on $ \triangle ABC$ with cevians AX and AY: $ AB^2(n + 1)/2 + AC^2(n - 1)/2 = AX^2n + (n + 1)(n - 1)n/4$ $ AB^2(n - 1)/2 + AC^2(n + 1)/2 = AY^2n + (n + 1)(n - 1)n/4$ Adding the two, and using that $ AB^2 + AC^2 = n^2$ by Pythag, we have $ AX^2 + AY^2 - 1 = (n^2 - 1)/2$ Which is what we set out to prove.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IMO1960Problem3 Vo Duc Dien

Consider a point $D$ on $BC$ and name the lengths like in the figure. Let $H$ be the pedal from $D$ on $AC$. Then $\tan \Psi = \frac{m}{l} = \frac{n}{l} \tan \theta$. Let $d=\frac{a}{n}$ be be the length of a segment divided. Then we have \[ \tan \alpha = \frac{p-q}{1+pq}\]where $p = \frac{a+d}{a-d}\tan\theta$, $q = \frac{a-d}{a+d} \tan \theta$. \[ \tan \alpha = \frac{4ad \sin \theta \cos \theta}{a^2-d^2} = \frac{4dh}{a^2-d} = \frac{4nh}{(n^2-1)a}\]as desired.

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