Question 1044773
triangles {{{BDR}}} and {{{ACR}}} are similar; so, corresponding sides are proportional

than we have:

{{{BD/AC=BR/AR}}}......since {{{BD=5m}}} is parallel to {{{AC=3m}}}, and {{{BR=3m+AR}}} we will have


{{{5m/3m=(3m+AR)/AR}}}


{{{5cross(m)/3cross(m)=(3m+AR)/AR}}}


{{{5/3=(3m+AR)/AR}}}


{{{5*AR=(3m+AR)*3}}}


{{{5*AR=9m+3*AR}}}


{{{5*AR-3*AR=9m}}}


{{{2*AR=9m}}}


{{{AR=9m/2}}}


{{{AR=4.5m}}}