Let $ P$ be midpoint of $ AB.$ Parallel project trapezoid $ ABCD$ to isosceles trapezoid $ A'B'C'D'.$ Midpoints $ P, M$ of $ AB, CD$ go to midpoints $ P', M'$ of $ A'B', C'D'$ and intersections $ Q \equiv DP \cap AC, N \equiv AM \cap BD$ to intersections $ Q' \equiv D'P' \cap A'C', N' \equiv A'M' \cap B'D'.$ By symmetry, $ Q'N' \parallel A'D'$ $ \Longrightarrow$ $ QN \parallel AD.$ Undefine point $ K$ and let $ K \equiv AC \cap BD.$ $ DC = DB$ $ \Longrightarrow$ $ \angle DBC = \angle BCD = 72^\circ.$ $ BC \parallel AD$ $ \Longrightarrow$ $ \angle BDA = \angle DBC = 72^\circ$ and $ \angle CDA = 180^\circ - \angle BCD = 108^\circ.$ $ DA = DC$ $ \Longrightarrow$ $ \angle ACD = \angle DAC = 36^\circ.$ It follows that the centrally similar $ \triangle s AKD \sim \triangle CKB$ are isosceles with $ AD = AK, CB = CK.$ $ DB = DA$ $ \Longrightarrow$ $ DP \perp AB$ bisects $ \angle BDA = 72^\circ$ $ \Longrightarrow$ $ \angle CDK = \angle KDQ = \angle QDA = 36^\circ$ $ \Longrightarrow$ $ CK = QA.$ But $ QN \parallel AD \parallel BC$ $ \Longrightarrow$ $ BK = ND$ as well.

My solution,probably different from above solutions First I would like to prove a lemma before coming to the problem: Let $ABC$ be a $72-72-36$ triangle.Let $a$ denote the length of equal side and $b$ denote the length of the third side. Then,$a^2-b^2=ab$ Proof:Let $AB=AC$ Let $D$ be a point on $AC$ such that $BD=BC$.Therefore $\triangle ABC$ and $\triangle BCD$ are similar.Also $BD$ is the angle bisector of $B$.So by angle bisector property we get that $CD=\frac {ba}{a+b}$Also since $ABC$ and $BCD$ are similar triangles we get that $AB\times CD=BC\times BC\Rightarrow a\times \frac {ab}{a+b}=b^2\Rightarrow a^2-b^2-ab=0$ Coming back to the main problem, Let a parallel to $AB$ through $C$ intersect $BD$ at point $X$ and $AD$ at point $T$ Claim:$N=X$. Proof:We observe that $BATC$ is a parallelogram.$\angle DCB=\angle DBC=\angle BCA$ and $\angle DAB=\angle TAB=\angle TCB=\angle XCB$ Thus triangles $DBA$ and $BCX$ are similar.Therefore,$AB\times BC=CX\times AD$...........(1) By the lemma discussed above and since $\triangle BCD$ is $36-72-72$,we get that $BC^2=CD^2+BC\times CD$.So rearranging the terms we get that $\frac{CD}{BC}-\frac{BC}{CD}=1$..................(2) Now to prove that $A,X$ and $M$ collinear,we use converse of Menelaus's theorem, We just need to prove that $TX\times AD$=$XC\times AT$ $XC\times AT=BC\times XC$. by (2) and (1) and by the fact $AD=CD$,we get $\frac {BC}{AD}=\frac {AB}{XC}-1=\frac {CT}{XC}-1=\frac {TX}{XC}$,So we get $A,X$ and $M$ are collinear as desired.So $X=N$. Also we observe that $\triangle AKD\equiv \triangle DBC$,So we get that $BC=DK$,also from previous claims $BC=BN$ So we get that $BK$=$ND$