《2025年启东中学作业本八年级数学上册苏科版徐州专版》

23.(12分)(东台期中)如图①,在长方形$ABCD$中,$AB= 5$,$AD= 12$,$E为AD$边上一点,$DE= 4$,动点$P从点B$出发,沿$B\to C\to D$以每秒2个单位长度的速度做匀速运动,设运动时间为$t(s)$.
(1)当$t$为何值时,$\triangle ABP与\triangle CDE$全等？
解:∵四边形ABCD是长方形，
∴AB = CD,∠B = ∠D = 90°.
当点P在BC边上,且BP = DE = 4时,△ABP≌△CDE;∵BP = 2t,∴2t = 4,解得t = 2.
当点P在CD边上，若点P与点C重合，满足AB = CD,∠B = ∠D = 90°,此时BP＞DE,△ABP与△CDE不全等,
若点P与点D重合,满足AB = CD,∠BAD = ∠D = 90°,此时AP＞DE,△BAP与△CDE不全等
综上所述，当t = 2时,△ABP与△CDE全等
(2)如图②,$EF为\triangle AEP$的高,当点$P在BC$边上运动时,$EF$的最小值为______.
$\frac{40}{13}$
(3)当点$P在EC$的垂直平分线上时,求$t$的值.
解:设EC的垂直平分线为直线MN.
如答图①,若点P在BC边上，且在直线MN上，连接PE,作PG⊥AD于点G,则PE = PC = 12 - 2t,∠PGE = 90°∵AD//BC,PG⊥AD,CD⊥AD,∴PG = CD = 5.
同理AG = BP = 2t,GE = 8 - 2t.
在Rt△EPG中,GE² + PG² = PE²,
即(8 - 2t)² + 5² = (12 - 2t)²,解得t = $\frac{55}{16}$.
如答图②,若点P在CD边上，且在直线MN上,连接PE,则PE = PC = 2t - 12.
∵DE² + PD² = PE²,且PD = 12 + 5 - 2t = 17 - 2t,
∴4² + (17 - 2t)² = (2t - 12)²,解得t = $\frac{161}{20}$.
综上所述，t的值为$\frac{55}{16}$或$\frac{161}{20}$.

答案：
(1)解:
∵四边形ABCD是长方形，
∴AB = CD,∠B = ∠D = 90°.
当点P在BC边上,且BP = DE = 4时,△ABP≌△CDE;
∵BP = 2t,
∴2t = 4,解得t = 2.
当点P在CD边上，若点P与点C重合，满足AB = CD,∠B = ∠D = 90°,此时BP＞DE,△ABP与△CDE不全等,
若点P与点D重合,满足AB = CD,∠BAD = ∠D = 90°,此时AP＞DE,△BAP与△CDE不全等
综上所述，当t = 2时,△ABP与△CDE全等
(2)$\frac{40}{13}$
(3)解:设EC的垂直平分线为直线MN.
如答图①,若点P在BC边上，且在直线MN上，连接PE,作PG⊥AD于点G,则PE = PC = 12 - 2t,∠PGE = 90°
∵AD//BC,PG⊥AD,CD⊥AD,
∴PG = CD = 5.
同理AG = BP = 2t,GE = 8 - 2t.
在Rt△EPG中,GE² + PG² = PE²,
即(8 - 2t)² + 5² = (12 - 2t)²,解得t = $\frac{55}{16}$.
如答图②,若点P在CD边上，且在直线MN上,连接PE,则PE = PC = 2t - 12.
∵DE² + PD² = PE²,且PD = 12 + 5 - 2t = 17 - 2t,
∴4² + (17 - 2t)² = (2t - 12)²,解得t = $\frac{161}{20}$.
综上所述，t的值为$\frac{55}{16}$或$\frac{161}{20}$.