Proving $\frac mn > \frac ba$ in $\triangle ABC$

Question

In the triangle $\triangle ABC$, assume $b > a$, prove $\dfrac mn > \dfrac ba$.

I solved this problem as follow,

Let $CH = h$ we have $b^2 = h^2 + m^2$ and $a^2 = n^2 + h^2$. Proving $\dfrac mn> \dfrac ba$ is equivalent to prove $\dfrac {m^2}{n^2} > \dfrac{b^2}{a^2}$. So we should prove $\dfrac{m^2}{n^2} > \dfrac{m^2 + h^2}{n^2+h^2}$ which is not a hard task for example by using the methods provided here.

I'm looking for other approaches to solve this problem especially with less algebraic manipulations and more geometry involved.

Answer 1

This is equivalent $\frac{m}{b} > \frac{n}{a}$. It means you can compare angles by the base, because these are their cosines. As measure of angle rises with length of opposite side and cosine is decreasing, so left angle is smaller than right one - $b>a$ so cosine is bigger of smaller angle. CBDO

Answer 2

Using that $\alpha = \angle CAH$ and $\beta = \angle CBH$ are both acute (so the $\sin$ and $\cos$ of both angles are positive), that $\sin^{2}x + \cos^{2}x = 1$, and all side lengths are positive, we get

$$a \lt b \;\to\; \frac{h}{b} \lt \frac{h}{a} \;\to\; \sin\alpha \lt \sin\beta \;\to\; \cos\alpha \gt \cos\beta \;\to\; \frac{m}{b} \gt \frac{n}{a} \;\to\; \frac{m}{n} \gt \frac{b}{a}$$