1. 如图6,在$□ ABCD$中,$AE⊥ BC,AF⊥ CD$,垂足分别为$E,F$,且$BE=DF$.
(1)求证:$□ ABCD$是菱形;

(2)若$AB=5,AC=6$,求$□ ABCD$的面积.

1. (1)略 (2)解:连结$BD$交$AC$于$O$,$\because$四边形$ABCD$是菱形,$\therefore AC\bot BD$,$AC$=6,$AO$=$OC$=$\dfrac{1}{2}AC$=$\dfrac{1}{2}$×6=3,$\because AB$=5,$AO$=3,$\therefore BO$=$\sqrt{AB^{2}-AO^{2}}$=$\sqrt{5^{2}-3^{2}}$=4,$\therefore BD$=2$BO$=8,$\therefore S_{□ ABCD}$=$\dfrac{1}{2}$×$AC$×$BD$=24

2. 如图7,在四边形$ABCD$中,$AB=AD,CB=CD,AB// CD$.
(1)求证:四边形$ABCD$是菱形;
(2)*点$E$是$CD$上一点,$BE$交$AC$于点$F$,连接$DF$. 若$BE⊥ CD$,求证:$∠ EFD=∠ BCD$.

2. 提示:(1)利用SSS证$△ ABC≌△ ADC$,得到$∠ BAC=∠ DAC$,又$\because AB// CD$,
$\therefore ∠ DCA=∠ BAC$,$\therefore ∠ DCA=∠ DAC$,$\therefore CD=AD$,又$\because AB=AD$,$CB=CD$,$\therefore AB=CB=AD=CD$,
$\therefore$四边形$ABCD$是菱形 (2)先证$△ CFB≌△ CFD$(SAS),从而得到$∠ CBF=∠ CDF$,又$\because BE\bot CD$,
$\therefore ∠ BED=∠ BEC=90°$,$\therefore ∠ EFD+∠ CDF=∠ BCD+∠ CBF=90°$,$\therefore ∠ EFD=∠ BCD$