11. 如图,矩形 $ABCD$ 的对角线相交于点 $O$,分别过点 $A$,$D$ 作 $AE// BD$,$DE// AC$ 交于点 $E$.求证:四边形 $AODE$ 是菱形.

答案
11. 证明:$\because AE// BD,DE// AC$,
$\therefore$四边形$AODE$是平行四边形,
$\because$四边形$ABCD$是矩形,$\therefore OA=OD$,
$\therefore$平行四边形$AODE$是菱形.
$\therefore$四边形$AODE$是平行四边形,
$\because$四边形$ABCD$是矩形,$\therefore OA=OD$,
$\therefore$平行四边形$AODE$是菱形.
12. 如图,$△ ABC$ 中,$AB = AC$,$AD$ 是 $△ ABC$ 外角的平分线,$∠ BAC=∠ ACD$.
(1)求证:$△ ABC≌△ CDA$.
(2)若 $∠ B = 60°$,求证:四边形 $ABCD$ 是菱形.

(1)求证:$△ ABC≌△ CDA$.
(2)若 $∠ B = 60°$,求证:四边形 $ABCD$ 是菱形.
答案
12. 证明:(1)$\because AB=AC$,$\therefore ∠ B=∠ ACB$,
$\because ∠ FAC=∠ B+∠ ACB=2∠ ACB$,$AD$平分$∠ FAC$,
$\therefore ∠ FAC=2∠ CAD$,且$∠ CAD=∠ ACB$,
$\because$在$△ ABC$和$△ CDA$中,$∠ BAC=∠ ACD$,
$AC=CA$,$∠ ACB=∠ CAD$,
$\therefore △ ABC≌△ CDA(ASA)$.
(2)$\because ∠ FAC=2∠ ACB$,$∠ FAC=2∠ DAC$,
$\therefore ∠ DAC=∠ ACB$,
$\therefore AD// BC$,
$\because ∠ BAC=∠ ACD$,
$\therefore AB// CD$,
$\therefore$四边形$ABCD$是平行四边形,
$\because ∠ B=60°$,$AB=AC$,
$\therefore △ ABC$是等边三角形,$\therefore AB=BC$,
$\therefore$平行四边形$ABCD$是菱形.
$\because ∠ FAC=∠ B+∠ ACB=2∠ ACB$,$AD$平分$∠ FAC$,
$\therefore ∠ FAC=2∠ CAD$,且$∠ CAD=∠ ACB$,
$\because$在$△ ABC$和$△ CDA$中,$∠ BAC=∠ ACD$,
$AC=CA$,$∠ ACB=∠ CAD$,
$\therefore △ ABC≌△ CDA(ASA)$.
(2)$\because ∠ FAC=2∠ ACB$,$∠ FAC=2∠ DAC$,
$\therefore ∠ DAC=∠ ACB$,
$\therefore AD// BC$,
$\because ∠ BAC=∠ ACD$,
$\therefore AB// CD$,
$\therefore$四边形$ABCD$是平行四边形,
$\because ∠ B=60°$,$AB=AC$,
$\therefore △ ABC$是等边三角形,$\therefore AB=BC$,
$\therefore$平行四边形$ABCD$是菱形.
13. 如图所示,在 $□ ABCD$ 中,$E$,$F$ 分别为边 $AB$,$CD$ 的中点,$BD$ 是对角线,过点 $A$ 作 $AG// DB$ 交 $CB$ 的延长线于点 $G$.
(1)求证:$DE// BF$.
(2)若 $∠ G = 90°$,求证:四边形 $DEBF$ 是菱形.

(1)求证:$DE// BF$.
(2)若 $∠ G = 90°$,求证:四边形 $DEBF$ 是菱形.
答案
13. 证明:(1)$\because$四边形$ABCD$是平行四边形,
$\therefore AB// CD$,$AB=CD$,
$\because$点$E$,$F$分别是$AB$,$CD$的中点,
$\therefore BE=\frac{1}{2}AB$,$DF=\frac{1}{2}CD$,
$\therefore BE=DF$,$BE// DF$,
$\therefore$四边形$DEBF$是平行四边形,$\therefore DE// BF$.
(2)$\because ∠ G=90°$,$AG// BD$,$AD// BG$,
$\therefore$四边形$AGBD$是矩形,
$\therefore ∠ ADB=90°$,
在$\mathrm{Rt}△ ADB$中,$\because E$为$AB$的中点,
$\therefore AE=BE=DE$,
$\because$四边形$DEBF$是平行四边形,
$\therefore$四边形$DEBF$是菱形.
$\therefore AB// CD$,$AB=CD$,
$\because$点$E$,$F$分别是$AB$,$CD$的中点,
$\therefore BE=\frac{1}{2}AB$,$DF=\frac{1}{2}CD$,
$\therefore BE=DF$,$BE// DF$,
$\therefore$四边形$DEBF$是平行四边形,$\therefore DE// BF$.
(2)$\because ∠ G=90°$,$AG// BD$,$AD// BG$,
$\therefore$四边形$AGBD$是矩形,
$\therefore ∠ ADB=90°$,
在$\mathrm{Rt}△ ADB$中,$\because E$为$AB$的中点,
$\therefore AE=BE=DE$,
$\because$四边形$DEBF$是平行四边形,
$\therefore$四边形$DEBF$是菱形.
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