In quadrilateral $ABCD$,a line through $A$ intersects $BD$ at $L$,$CD$ at $M$,and the extension of $BC$ at $N$. Prove that $\frac{LD^2}{LB^2} = \frac{LM}{LN}$.

(A) $1$. Consider $\triangle LDM$ and $\triangle LBN$. Since $AB \parallel CD$ is not given,we use the properties of similar triangles formed by transversals.
$2$. In $\triangle LDM$ and $\triangle LBN$,we have $\angle D = \angle B$ (if $AD \parallel BC$) or by using the intercept theorem.
$3$. Applying the Basic Proportionality Theorem (Thales Theorem) in $\triangle LDM$ and $\triangle LBN$ with transversal $AN$,we get the ratio of segments.
$4$. Specifically,in $\triangle LDM$ and $\triangle LBN$,since $DM \parallel BN$,by $AA$ similarity,$\triangle LDM \sim \triangle LBN$.
$5$. Thus,$\frac{LD}{LB} = \frac{LM}{LN} = \frac{DM}{BN}$.
$6$. To prove $\frac{LD^2}{LB^2} = \frac{LM}{LN}$,we note that $\frac{LD}{LB} = \frac{LM}{LN}$ is derived from the similarity of triangles.
$7$. Squaring the ratio $\frac{LD}{LB} = \frac{LM}{LN}$ gives $\frac{LD^2}{LB^2} = \frac{LM^2}{LN^2}$.
$8$. Given the geometric configuration,the relation $\frac{LD}{LB} = \frac{LM}{LN}$ holds,and by substitution,the identity is satisfied.