Locating a point of simultaneous arrival Please consider the following image. Suppose there are two persons, person $1$ walks from $A$ to $B$, while person $2$ walks from $A$ to $D$ to $B$ such that person $1$ reaches at $B$ before person $2$. Also, the speed of person $1$ is faster than person $2$. Given these conditions is it possible for person $1$ and person $2$ to reach at point $C$ simultaneously, such that person $1$ goes $A$ to $B$ to $C$, while person $2$ goes $A$ to $D$ to $C$. 

What have I tried.
Let's say:
$$
\alpha=\frac{v_2}{v_1}
$$
where


*

*$v_1$ is the speed of person 1 

*$v_2$ is the speed of person 2


Assuming that both person meet at $C$ simultaneously:
$$
a_1+a_3=\alpha(b_1+b_2)
$$
Also since person $1$ reaches at $B$ before person $2$:
$$
a_1+a_2>\alpha\cdot b_1
$$
From triangle DBC:
$$
a_3+b_2>a_2 
$$ 
and (putting this in previous equation)
$$
a_1+a_3+b_2>\alpha\cdot b_1
$$
By applying the first equation:
$$
\alpha(b_1+b_2)+b_2>\alpha\cdot b_1
$$
So we get:
$$
(\alpha+1)b_2>0
$$
So I am tring to prove that such a point is not possible through contradiction but not getting a successful result.
 A: Your intuition is correct. Let me give you an informal explanation before diving into equations: since person 1 is clearly faster, they would definitely win if in the "trial run" were extended so that both have to walk from B to C afterward. But it's even faster for person 1 to travel from D to C (as in the "real run") than it is to pass through B first (as in the extended trial).

Using distance = rate * time, we observe that the trial run (i.e. getting from A to B) tells us the following information, applying the triangle inequality:
$$v_1\tau_1=a_1+a_2>b_1=v_2\tau_2, \qquad\text{and}\qquad \tau_1<\tau_2.$$
Here, $\tau_i$ is the time it took person $i$ to go from $A$ to $B$, and $v_i$ is the speed of person $i$.
In particular, this means that $v_1>v_2$, and so 
$$\frac{b_2}{v_1}<\frac{b_2}{v_2}.$$ 
Converting the $\tau$ inequality to the other variables we get:
$$\frac{a_1+a_2}{v_1}<\frac{b_1}{v_2}$$
Adding these together gives:
$$\frac{a_1+a_2+b_2}{v_1}<\frac{b_1+b_2}{v_2}$$
The triangle inequality also implies that $a_3<a_2+b_2$, and so
$$ t_1 = \frac{a_1+a_3}{v_1}< \frac{b_1+b_2}{v_2}=t_2$$
If we assume that the people walk at the same speed during the real run from A to C, as they did inn the trial run, then we've shown that person $1$ gets to C first.
