Given that R is interior to triangle BAD, prove $BR+DR<BA+AD$

Question

I need help with this proof for my math class!

Answer 1

Suppose A($A_x$, $A_y$), B(0, 0), D($D_x$, 0) and R(c, e). $${\left( c \lt A_x \right)} \rightarrow {\left( c \gt \frac{A_x \times e}{A_y} \right)} \land {\left( e \lt \frac{A_y \times c}{A_x} \right)}$$, $${\left( c \gt A_x \right)} \rightarrow {\left( c \lt D_x - \frac{{\left( D_x - A_x \right)} \times e}{A_y} \right)} \land {\left( e \lt A_y - \frac{A_y \times {\left( c + A_x \right)}}{A_x} \right)}$$, ${\left( c = A_x \right)} \rightarrow {\left( e \lt A_y \right)}$, $BR = {\sqrt{c^2 + e^2}}$, $DR = {\sqrt{{\left(D_x - c\right)}^2 + e^2}}$, $BA = {\sqrt{A_x^2 + A_y^2}}$, $AD = {\sqrt{{\left( D_x - A_x \right)}^2 + A_y^2}}$, As all variables are positive, check $BA + AD - BR - RD \gt 0$, this process can take long time: ●If c = $A_x$ then e < $A_y$. Therefore, BA > BR and AD > RD. ●If $c \neq A_x$ then recall ${\left( a - b \right)}^2 = {\left( b - a \right)}^2$. Furthermore, BA + AD > BR + RD. Finally BA + AD > BR + RD.