6. 如图,在梯形$ABCD$中,$AD// BC(BC＞AD)$,$E,F$分别是对角线$BD$,$AC$的中点.求证:
$EF=\dfrac{1}{2}(BC-AD)$.

6.如图,连接AE并延长,交BC于点G.

$\because AD// BC$,
$\therefore ∠ ADE=∠ GBE$,$∠ EAD=∠ EGB$.
又$E$为$BD$中点,$\therefore DE=BE$,$\therefore △ AED≌△ GEB(\mathrm{AAS})$,$\therefore BG=AD$,$AE=EG$.
在$△ AGC$中,$EF$为中位线,
$\therefore EF=\dfrac{1}{2}GC=\dfrac{1}{2}(BC-BG)=\dfrac{1}{2}(BC-AD)$,即$EF=\dfrac{1}{2}(BC-AD)$.

7. 如图(1),在正方形ABCD中,E,F分别是AD,DC上的点,且$AF⊥BE$.
(1)求证:$AF=BE$.
(2)如图(2),在正方形ABCD中,M,N,P,Q分别是AB,BC,CD,DA上的点,且$MP⊥NQ$. MP与NQ是否相等？试说明理由.

7.(1)设AF与BE交于点G.
$\because$ 四边形ABCD是正方形,
$\therefore AB=AD$,$∠ BAD=∠ D=90°$,
$\therefore ∠ FAD+∠ AFD=90°$.
$\because AF⊥ BE$,$\therefore ∠ AGE=90°$,
$\therefore$ 在$\mathrm{Rt}△ AGE$中,$∠ FAD+∠ AEG=90°$,
$\therefore ∠ AFD=∠ AEG$.
在$△ DAF$和$△ ABE$中,
$\begin{cases}∠ D=∠ BAE,\\∠ AFD=∠ BEA,\\DA=AB,\end{cases}$
$\therefore △ DAF≌△ ABE(\mathrm{AAS})$,$\therefore AF=BE$.
(2)如图,过点A作$AF// MP$交CD于点F,过点B作$BE// NQ$交AD于点E,得到四边形AFPM,BEQN均为平行四边形,
$\therefore AF=MP$,$BE=NQ$.
$\because MP⊥ NQ$,$AF// MP$,
$\therefore AF⊥ NQ$.
又$BE// NQ$,$\therefore AF⊥ BE$.
由(1),得$AF=BE$,
$\therefore MP=NQ$.

8. 如图,在$△ ABC$中,$∠ ABC=90°$,$BD$是$△ ABC$的中线,$F$是$BD$的中点,连接$CF$并延长到点$E$,使$FE=CF$,连接$BE$,$AE$.
(1)求证:四边形$AEBD$是菱形;
(2)若$BC=8$,$BE=5$,求菱形$AEBD$的面积.

8.(1)$\because F$是BD的中点,
$\therefore DF=BF$.
$\because CF=EF$,$∠ CFD=∠ EFB$,
$\therefore △ CDF≌△ EBF(\mathrm{SAS})$,
$\therefore CD=BE$,$∠ FCD=∠ FEB$,
$\therefore BE// CD$.
$\because ∠ ABC=90°$,$BD$是$△ ABC$的中线,
$\therefore BD=\dfrac{1}{2}AC=AD=CD$,
$\therefore BE=CD=AD$,
$\therefore$ 四边形AEBD是平行四边形.
$\because BD=AD$,
$\therefore$ 四边形AEBD是菱形.
(2)菱形AEBD的面积为24.