How to prove that $v_3t_1=t_2v_1$ for projectile motion I have the following equation before me:
Let $\alpha$, $\alpha-\beta$ and $\alpha-2\beta$ be the angles made with the horizontal by a projectile at three points $A$, $B$ and $C$ respectively where velocities are $v_1$,$v_2$ &$v_3$. Let $t_1$ and $t_2$ be the times required to describe arcs $AB$ and $BC$ respectively.
I have to prove that $v_3t_1=t_2v_1$.
To begin with, I have the following set of equations which is obtained by considering velocity in $x$ direction.
$v_1\cos(\alpha)=v_2\cos(\alpha-\beta)=v_3\cos(\alpha-2\beta)$
By considering velocity in $y$ direction, I get the following equation:
$v_2\sin(\alpha-\beta)=v_1\sin(\alpha)-gt_1$
$v_3\sin(\alpha-2\beta)=v_2\sin(\alpha-\beta)-gt_2$
These two equations can be combined into one given below:
$\dfrac{t_1}{t_2}=\dfrac{v_1\sin(\alpha)-v_2\sin(\alpha-\beta)}{v_2\sin(\alpha-\beta)-v_3\sin(\alpha-2\beta)}$
How can I get the required result by application of these equations? That's where I am stuck. Please suggest how to proceed further.

Take your last equation. In the numerator write $v_2$ in terms of $v_1$, while in the denominator you write $v_2$ in terms of $v_3$. You get these relationships from the velocities you wrote in the horizontal direction.
$$\frac{t_1}{t_2}=\frac{v_1\sin\alpha-v_1\frac{\cos\alpha}{\cos(\alpha-\beta)}\sin(\alpha-\beta)}{v_3\frac{\cos(\alpha-2\beta)}{\cos(\alpha-\beta)}\sin(\alpha-\beta)-v_3\sin(\alpha-2\beta)}$$
Now take $v_1/v_3$ in front, and use common denominator for both numerator and denominator (notice that it's the same)
$$\frac{t_1}{t_2}=\frac{v_1}{v_3}\frac{\sin\alpha\cos(\alpha-\beta)-\cos\alpha\sin(\alpha-\beta)}{\sin(\alpha-\beta)\cos(\alpha-2\beta)-\cos(\alpha-\beta)\sin(\alpha-2\beta)}$$
Now use $\sin(x-y)=\sin x\cos y-\cos x\sin y$ and you see
$$\frac{t_1}{t_2}=\frac{v_1}{v_3}\frac{\sin(\alpha-(\alpha-\beta))}{\sin(\alpha-\beta-(\alpha-2\beta))}=\frac{v_1}{v_2}\frac{\sin\beta}{\sin\beta}=\frac{v_1}{v_3}$$
