In a triangle of base $a$ the ratio of the other two sides is $r(<1)$. Show that the altitude of the triangle is less than or equal to $\frac{ar}{1-r^2}$.
I noticed that for a triangle of base $a$, for any given altitude $A$, it is always possible to construct a triangle in which the other two sides are in ratio $r(<1)$ by considering an isosceles triangle with base $a$ and altitude $A$ where the two equal sides are other than the base, and by shifting its apex perpendicular to the altitude to the extent that it is required to obtain a ratio of $r(<1)$ between the other two sides. If my observation is right - and if it is not, I would appreciate it if somebody could explain why - how can be there be an upper bound on the altitude?
Let sides of length $a, b, c=br$ correspond to angle $A, B ,C$ and let the height be $h$.
Using the area of triangle and sine law $b\sin(A)=a\sin(B)$,
$${1\over 2}ah = {1\over 2}b^2r\sin(A) = {1\over 2}a^2r{\sin^2(B)\over \sin(A)}$$
Simplify, $$h = ar{\sin^2(B)\over \sin(A)}$$
Note that
$$\sin(A)=\sin(B+C)\geq\sin(B+C)\sin(B-C)=\sin^2(B)-\sin^2(C)$$
the last equality is a known fact where proof can be found here
Also since $r={\sin(C)\over \sin(B)}$, therefore
$$h = ar{\sin^2(B)\over \sin(A)} \leq ar{\sin^2(B)\over \sin^2(B)-\sin^2(C)}=ar{1\over 1-({\sin^2(C)\over \sin^2(B)})^2}=ar{1 \over 1-r^2}$$