Let $u$ and $v$ be the distances of an arbitrary point on the side $AB$ of an acute-angled triangle $ABC$ to its sides $BC$ and $AC$. Let $h_a$ and $h_b$ be the lengths of the altitudes from its vertices $A$ and $B$ respectively. Prove that
$$$$ $min$ {$h_a,h_b$} $\leq$ $u+v$ $\leq$ $max$ {$h_a,h_b$}
$$$$
I was able to prove the first part of the question like this: Let the arbitrary point be X.
WLOG, let $h_a\geq h_b$. Then$$\frac{AX}{XB}={h_a\over v}={u\over h_b}$$
$$h_ah_b=uv$$
$$u+v\geq2\sqrt{uv}=2\sqrt{h_ah_b}\geq2h_b$$
$$Or, u+v\geq min\{h_a,h_b\}$$
But now I'm stuck in the second part. $$$$
Edit: The above proof is INCORRECT; the correct proof is given below in @player3236 's answer.
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1 Answer
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Your idea (of using similar triangles) is correct but the formula is wrong. We should have:
$$\frac {AX}{AB} = \frac {v}{h_b}, \quad \frac {BX}{AB} = \frac {u}{h_a}$$
Hence
$$u+v = h_a\frac {BX}{AB} + h_b\frac {AX}{AB}$$
Therefore
$$\min\{h_a,h_b\}\frac {BX}{AB} + \min\{h_a,h_b\}\frac {AX}{AB} \le h_a\frac {BX}{AB} + h_b\frac {AX}{AB} \le \max\{h_a,h_b\}\frac {BX}{AB} + \max\{h_a,h_b\}\frac {AX}{AB}$$
which simplifies to
$$\min\{h_a,h_b\} \le u+v \le \max\{h_a,h_b\}$$
