《2025年学霸提优大试卷八年级数学上册苏科版》

24.(8分)如图,在$\triangle ABC$中,$BD \perp AC$于点D,$CE \perp AB$于点E,点M,N分别是BC,DE的中点.
(1)求证:$MN \perp DE$;
(2)若$\angle ECB + \angle DBC = 45^{\circ},DE = 10$,求MN的长.

答案：

(1)如图,连接EM,DM,$\because BD\perp AC$,$CE\perp AB$,$\therefore \angle BDC=\angle BEC=90^{\circ}$.$\because$ 在$\text{Rt}\triangle DBC$和$\text{Rt}\triangle EBC$中,M是斜边BC的中点,$\therefore DM=\frac{1}{2}BC$,$EM=\frac{1}{2}BC$,$\therefore DM=EM$.$\because N$是DE的中点,$\therefore MN\perp DE$.
(2)$\because DM=\frac{1}{2}BC=BM$,$\therefore \angle DBM=\angle BDM$,同理$\angle MEC=\angle MCE$.$\because \angle ECB+\angle DBC=45^{\circ}$,$\therefore \angle EMB+\angle DMC=2(\angle ECB+\angle DBC)=90^{\circ}$,$\therefore \angle EMD=90^{\circ}$.$\because N$是DE的中点,$DE=10$,$\therefore MN=\frac{1}{2}DE=5$.

25.(10分)如图①,已知长方形ABCD,$AB = CD = 2,BC = AD = 3,\angle A = \angle B = \angle C = \angle D = 90^{\circ}$,E为CD边的中点,P为长方形ABCD边上的动点,动点P从点A出发,沿着$A \to B \to C \to E$运动到点E停止,设点P经过的路程为x,$\triangle APE$的面积为y.
(1)当$x = 1$时,$y = $;当$x = 5.5$时,$y = $.
(2)如图②,当点P在边BC上时,用含x的代数式表示y.
(3)当P在线段BC上运动时,是否存在点P,使得$\triangle APE$的周长最小?若存在,求出此时$\angle PAD$的度数;若不存在,请说明理由.

答案：

(1)1.5 0.75 【解析】当$x=1$时,点P在AB上,$AP=1$,$AD=3$,$\therefore y=\frac{1}{2}× 1× 3=1.5$;当$x=5.5$时,点P在CE上,$EP=2 + 3+1 - 5.5=0.5$,$AD=3$,$\therefore y=\frac{1}{2}× 0.5× 3=0.75$.
(2)当点P在BC边上时,$BP=x - 2$,$CP=5 - x$,$\therefore y=2× 3-\frac{1}{2}× 2× (x - 2)-\frac{1}{2}× 1× (5 - x)-\frac{1}{2}× 1× 3=-\frac{1}{2}x + 4(2\leqslant x\leqslant 5)$.
(3)存在.如图,作点E关于BC所在直线的对称点$E'$,连接$AE'$,交BC于点P,此时$\triangle APE$的周长最小.$\because EC=CE'$,且$PC\perp EE'$,$\therefore PE=PE'$,$\therefore AP+PE=AE'$.$\because AE$为定值,$\therefore$ 此时$\triangle APE$的周长最小.在$\text{Rt}\triangle ADE'$中,$\because AD=DE'=3$,$\angle D=90^{\circ}$,$\therefore \triangle ADE'$是等腰直角三角形,$\therefore \angle PAD=45^{\circ}$.