Another Quadrilateral Problem

First we construct two lines, one parallel to $AD$ passing on $B$ and the line $BD$ :

Now we notice that $\angle BEC = \angle ADB = \angle BAD$ and $\angle BCE = \angle ADB\rightarrow \angle CBE = \angle ADB$ . With that, $\triangle ABD$ and $\triangle BCE$ are similars. Now we can make the propostion: $\frac {CE}{6} = \frac {8}{10}\rightarrow CE = 4.8$ . Finally we have that $AB=DE=8$ because $\angle BAD = \angle ADE$ . Our answer will be: $DE + CE = 8 + 4.8 = \boxed{12.8}$

Let $\angle BAD=X$ and $\angle ABD=Y$ . Draw line $l$ parallel to side $AD$ passing through $B$ . Construct a point $E$ such that quadrilateral $ABED$ is an isoceles trapezoid .From here we have two cases:( $i$ ) $BE$ is outside $ABCD$ and ( $ii$ ) $BE$ lies inside $CD$ but $i$ does not work because because one will find from angle chase that either of angle $\angle EBC$ or $\angle CBD$ is negative. Hence, $BE$ is inside $ABCD$ .From angle chase one will find that $BE$ bisects $\angle B$ . By internal angle bisector theorem we find $EC=4.8$ . Therefore, $CD=12.8$ .

well i used a bit of sine rule and the constructions which mietantei conan has done