《2025年实验班提优大考卷八年级数学上册苏科版》

26. 如图,在△ABC中,AB= AC= 10,AD是边BC上的高,点E在边BC上,连接AE.
(1)当AD= 6时,
①求△ABC的面积；
②若AE平分∠BAD,求CE的长.
(2)探求三条线段AE,BE,CE之间的等量关系.

答案：
(1)①$\because$AD 是边 BC 上的高,$\therefore ∠ADB = 90^{\circ }$,$\therefore AD^{2}+BD^{2}=AB^{2}$.又$AB = 10$,$AD = 6$,$\therefore BD=\sqrt{10^{2}-6^{2}}=8$.$\because AB = AC$,AD 是边 BC 上的高,$\therefore BC = 2BD = 16$,$\therefore △ABC$的面积$=\frac{1}{2}BC\cdot AD=\frac{1}{2}×16×6 = 48$.②过点 E 作$EF⊥AB$于点 F.$\because$AE 平分$∠BAD$,$∠ADB = 90^{\circ }$,$\therefore EF = ED$.又$AE = AE$,$\therefore Rt△AEF\cong Rt△AED(HL)$,$\therefore AF = AD = 6$,$\therefore BF = 10 - 6 = 4$.设$EF = ED = x$,则$BE = 8 - x$,在$Rt△BEF$中,由勾股定理,得$x^{2}+4^{2}=(8 - x)^{2}$,解得$x = 3$,$\therefore CE = 8 + 3 = 11$.
(2)由勾股定理,得$AD^{2}+BD^{2}=AB^{2}$,$AD^{2}+DE^{2}=AE^{2}$,$\therefore AB^{2}-BD^{2}=AE^{2}-DE^{2}$.$\because AB = 10$,$\therefore AE^{2}+(BD^{2}-DE^{2})=100$.$\because AB = AC$,$AD⊥BC$,$\therefore BD = CD$,$\therefore AE^{2}+(BD^{2}-DE^{2})=AE^{2}+(BD + DE)(BD - DE)=AE^{2}+CE\cdot BE$,$\therefore AE^{2}+CE\cdot BE = 100$.