Prove AD= 3AB show exact steps to solve use algebraic terms for each step " Distributive property or Associative Multiplicative property" as of that

The question is to prove that
[tex]AD=3AB[/tex]From the diagram in the question, we can deduce that
[tex]AD=AB+BC+CD\text{ (TOTAL DIsTANCE FOR THE LINE)}[/tex]Also given from the question, we have
B is the midpoint of AC, that is
[tex]AB=BC(\text{ B bisects AC into two equal parts)}[/tex]Also given from the question, we have
C is the midpoint of BD, that is
[tex]BC=CD\text{ ( C bisects BD into two equal parts)}[/tex]Therefore,
since BC=CD ,and BC=AB then,
[tex]CD=AB\text{ (THE LINE IS DIVIDED INTO EQUAL SEGMENTS)}[/tex]By substituting the values of BC and CD in the total distance for the line , we will have
[tex]\begin{gathered} AD=AB+BC+CD \\ AD=AB+AB+AB \\ AD=3AB\text{ (AD IS 3 times the length of AB because the line is divided into equal lengths)} \\ \text{PROVED} \end{gathered}[/tex]