《2025年暑假大串联安徽人民出版社八年级数学北师大版》

1. 如图,A,B,C分别是线段$A_{1}B$,$B_{1}C$,$C_{1}A$的中点,若$\triangle ABC$的面积是1,那么$\triangle A_{1}B_{1}C_{1}$的面积是多少？

答案： 解:连接 $ AB_{1},BC_{1},CA_{1} $.
$\because AB = AA_{1}$,
$\therefore \triangle ABC$ 与 $\triangle AA_{1}C$ 的面积相等;
$\because AC = CC_{1}$,
$\therefore \triangle ABC$ 与 $\triangle BCC_{1}$ 的面积相等;
……
即 $\triangle ABC$ 与 $\triangle AA_{1}C,\triangle BCC_{1},\triangle A_{1}CC_{1},\triangle BB_{1}C_{1},\triangle ABB_{1},\triangle AA_{1}B_{1}$ 的面积相等,都等于 $ 1 $,
$\therefore \triangle A_{1}B_{1}C_{1}$ 的面积为 $ 7 $.

2. 如图,M是$\triangle ABC$的边BC的中点,AN平分$\angle BAC$,$BN\perp AN$于点N,延长BN交AC于点D,已知$AB = 10$,$BC = 15$,$MN = 3$。
(1)求证：$BN = DN$；
(2)求$\triangle ABC$的周长。

答案：
(1)证明:$\because AN$ 平分 $\angle BAC$,
$\therefore \angle BAN = \angle DAN$.
$\because BN \perp AN$,
$\therefore \angle ANB = \angle AND = 90^{\circ}$.
又 $\because AN = AN$,
$\therefore \triangle ABN \cong \triangle ADN$,
$\therefore BN = DN$.
(2)解:$\because$ 由
(1)得 $ BN = DN $,且 $ BM = CM,MN = 3 $,
$\therefore CD = 2MN = 6$.
$\because \triangle ABN \cong \triangle ADN,AB = 10$,
$\therefore AD = AB = 10$.
$\therefore \triangle ABC$ 的周长 $ = 10 + 10 + 6 + 15 = 41 $.

3. 如图,在$\triangle ABC$中,$\angle C = 90^{\circ}$,AD平分$\angle CAB$,交CB于点D,过点D作$DE\perp AB$于点E。
(1)求证：$\triangle ACD\cong\triangle AED$；
(2)若$\angle B = 30^{\circ}$,$CD = 1$,求BD的长。

答案：
(1)证明:$\because AD$ 平分 $\angle CAB$,
$\therefore \angle CAD = \angle EAD$.
$\because DE \perp AB,\angle C = 90^{\circ}$,
$\therefore \angle ACD = \angle AED = 90^{\circ}$.
又 $\because AD = AD$,
$\therefore \triangle ACD \cong \triangle AED(AAS)$.
(2)解:$\because \triangle ACD \cong \triangle AED$,
$\therefore DE = CD = 1$.
$\because \angle B = 30^{\circ},\angle DEB = 90^{\circ}$,
$\therefore BD = 2DE = 2$.

4. 如图,$\square ABCD$中,过A作$AM\perp BC$于M,交BD于E,过C作$CN\perp AD$于N,交BD于F,连接AF,CE。求证：四边形AECF为平行四边形。

答案： 证明:在 $\square ABCD$ 中,$ AB // CD,AB = CD,\angle ABC = \angle ADC $,
$\therefore \angle ABD = \angle CDB$.
又 $\because AM \perp BC,CN \perp AD$,
$\therefore \angle BAM = \angle DCN$,
$\therefore \triangle ABE \cong \triangle CDF$,
$\therefore AE = CF,\angle AEB = \angle CFD$,
$\therefore \angle AEF = \angle CFE$,
$\therefore AE // CF$,
$\therefore$ 四边形 $ AECF $ 为平行四边形.