Put $z=a$ then $|a-b|\geq |a-c|+|a-d|$ Put $z=b$ then $|b-a|\geq |b-c|+|b-d|$ Adding them we get that $2|a-b|\geq |a-c|+|a-d|+|b-c|+|b-d|$ $|a-c|+|b-c|\geq |a-b|$ and $|a-d|+|b-d|\geq |a-b|$ So this will be tru e when $c,d$ lies on the segment joining the complex numbers $a,b$ in the Argand plane. Now let $c=at+(1-t)b$ and $d=at'+(1-t')b$ where $t,t'\in (0,1)$ Now again putting $z=a$ we get that $|a-b|\geq (1-t)|a-b|+(1-t')|a-b|$ So $t+t'\geq 1$ Now putting $z=b$ we get that $|a-b|\geq t|a-b|+t'|a-b|$ So $t+t'=1$ hence done.

Generalization Let $k\in \left[ 0,\infty \right)$.To be considered the following complex and distinct $a,b,c,d$. Prove that the following affirmations are equivalent: i)For every$z\in \mathbb{C}$ the inequality takes place: $\left| z-a \right|+\left| z-b \right|+k\left| 2z-a-b \right|\ge \left| z-c \right|+\left| z-d \right|+k\left| 2z-c-d \right|$; ii)There is $t\in \left( 0,1 \right)$ so that $c=ta+\left( 1-t \right)b$ si $d=\left( 1-t \right)a+tb$