In $\Delta ABC$,$B-D-C$. The bisectors of $\angle ADB$ and $\angle ADC$ intersect $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$ respectively. Prove that $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$. From this,prove that $D$ is the midpoint of $\overline{BC}$ if $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$.

(N/A) $1$. In $\Delta ADB$,$DP$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$,which implies $AP = \frac{AD \times PB}{BD}$.
$2$. In $\Delta ADC$,$DQ$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AQ}{QC} = \frac{AD}{DC}$,which implies $AQ = \frac{AD \times QC}{DC}$.
$3$. Multiplying these two equations: $AP \times AQ = \frac{AD \times PB}{BD} \times \frac{AD \times QC}{DC} = \frac{AD^2 \times PB \times QC}{BD \times DC}$.
$4$. Rearranging gives $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$.
$5$. If $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$,then by the Basic Proportionality Theorem in $\Delta ABC$,$\frac{AP}{PB} = \frac{AQ}{QC}$.
$6$. From the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$ and $\frac{AQ}{QC} = \frac{AD}{DC}$.
$7$. Therefore,$\frac{AD}{BD} = \frac{AD}{DC}$,which implies $BD = DC$. Thus,$D$ is the midpoint of $\overline{BC}$.