It is known that: the common part BD of line AB and CD is 1 / 3AB = 1 / 5CD, the distance between the midpoint E and F of line AB and CD is 6cm, and the length of line AB and CD is calculated |_________ |_____ |____ |____ |_______ | A E D B F C

It is known that: the common part BD of line AB and CD is 1 / 3AB = 1 / 5CD, the distance between the midpoint E and F of line AB and CD is 6cm, and the length of line AB and CD is calculated |_________ |_____ |____ |____ |_______ | A E D B F C

BD=1/3AB
BE=1/2AB
So BD / be = 1 / 3 △ 1 / 2 = 2 / 3
So BD = 2 / 3 * be
BD=1/5CD
DF=1/2CD
So BD = 2 / 5 * DF
EF=BE+DF-BD
EF=BE+DF-2/3*BE=6
EF=BE+DF-2/5DF=6
That is 1 / 3bE + DF = 6
BE+3/5DF=6
Be = 3cm
DF=5cm
Then AB = 2be = 6cm
CD=2DF=10cm
Hope to help you