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17. (本小题8分)已知□ABCD的周长为40 cm，AB:BC = 2:3，求CD和AD的长.
答案：
解：$\because$四边形$ABCD$是平行四边形，
$\therefore AB = CD$，$AD = BC$. 设$AB = CD = 2a\ cm$，$AD = BC = 3a\ cm$，
又$\because$平行四边形$ABCD$的周长为$40\ cm$，
$\therefore 2(2a + 3a)= 40$，解得$a = 4$，
$\therefore CD = 8\ cm$，$AD = 12\ cm$.
$\therefore AB = CD$，$AD = BC$. 设$AB = CD = 2a\ cm$，$AD = BC = 3a\ cm$，
又$\because$平行四边形$ABCD$的周长为$40\ cm$，
$\therefore 2(2a + 3a)= 40$，解得$a = 4$，
$\therefore CD = 8\ cm$，$AD = 12\ cm$.
18. (本小题8分)如图，已知BD是△ABC的角平分线，点E，F分别在边AB，BC上，ED//BC，EF//AC. 求证：BE = CF.
答案：
证明：$\because ED// BC$，$EF// AC$，
$\therefore$四边形$EFCD$是平行四边形，$\therefore ED = CF$.
$\because BD$平分$\angle ABC$，$\therefore \angle EBD = \angle DBC$.
$\because ED// BC$，$\therefore \angle EDB = \angle DBC$，
$\therefore \angle EBD = \angle EDB$，$\therefore BE = ED$，$\therefore BE = CF$.
$\therefore$四边形$EFCD$是平行四边形，$\therefore ED = CF$.
$\because BD$平分$\angle ABC$，$\therefore \angle EBD = \angle DBC$.
$\because ED// BC$，$\therefore \angle EDB = \angle DBC$，
$\therefore \angle EBD = \angle EDB$，$\therefore BE = ED$，$\therefore BE = CF$.
19. (本小题9分)如图，四边形ABCD是平行四边形，AD = 12，AB = 13，BD⊥AD，求BC，CD及OB的长.
答案：
解：$\because$四边形$ABCD$是平行四边形，
$\therefore BC = AD = 12$，$CD = AB = 13$，$OB=\frac{1}{2}BD$.
$\because BD\perp AD$，$\therefore BD = \sqrt{AB^{2}-AD^{2}}=\sqrt{13^{2}-12^{2}} = 5$，
$\therefore OB=\frac{1}{2}BD=\frac{5}{2}$.
$\therefore BC = AD = 12$，$CD = AB = 13$，$OB=\frac{1}{2}BD$.
$\because BD\perp AD$，$\therefore BD = \sqrt{AB^{2}-AD^{2}}=\sqrt{13^{2}-12^{2}} = 5$，
$\therefore OB=\frac{1}{2}BD=\frac{5}{2}$.
20. (本小题10分)如图，在□ABCD中，对角线AC交BD于点O，四边形AODE是平行四边形. 求证：四边形ABOE是平行四边形.
答案：
解：$\because$在平行四边形$ABCD$中，
对角线$AC$交$BD$于点$O$，$\therefore OB = OD$.
又$\because$四边形$AODE$是平行四边形，
$\therefore AE// OD$，$AE = OD$，$\therefore AE// OB$，$AE = OB$，
$\therefore$四边形$ABOE$是平行四边形.
对角线$AC$交$BD$于点$O$，$\therefore OB = OD$.
又$\because$四边形$AODE$是平行四边形，
$\therefore AE// OD$，$AE = OD$，$\therefore AE// OB$，$AE = OB$，
$\therefore$四边形$ABOE$是平行四边形.
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