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電子課本網(wǎng) 第85頁(yè)

第85頁(yè)

信息發(fā)布者:
$m$
135°或45°
45
45
45
(1)證明:連接$CE,$因?yàn)?D$是$BC$的中點(diǎn),且$DE\perp BC,$所以$CE = BE。$又因?yàn)?BE^{2}-EA^{2}=AC^{2},$所以$CE^{2}-EA^{2}=AC^{2},$即$EA^{2}+AC^{2}=CE^{2},$根據(jù)勾股定理的逆定理可知$\triangle ACE$是直角三角形,所以$\angle A = 90^{\circ}。$
(2)解:因?yàn)?DE = 3,$$BD = 4,$所以$BE^{2}=DE^{2}+BD^{2}=3^{2}+4^{2}=25 = 5^{2},$所以$BE = CE = 5。$因?yàn)?BC = 2BD = 8,$在$Rt\triangle BAC$中,由勾股定理得$BC^{2}-BA^{2}=AC^{2},$即$64-(5 + AE)^{2}=25 - AE^{2},$
$\begin{aligned}64-(25 + 10AE+AE^{2})&=25 - AE^{2}\\64 - 25 - 10AE - AE^{2}&=25 - AE^{2}\\39 - 10AE - AE^{2}+AE^{2}&=25\\-10AE&=25 - 39\\-10AE&=-14\\AE&=\frac{7}{5}\end{aligned}$
證明:(2)連接$CE,$因?yàn)?\triangle ABC$繞頂點(diǎn)$B$按順時(shí)針?lè)较蛐D(zhuǎn)$60^{\circ},$得到$\triangle DBE,$所以$AC = DE,$$BC = BE,$$\angle CBE = 60^{\circ},$所以$\triangle BCE$是等邊三角形,所以$EC = BC,$$\angle BCE = 60^{\circ}。$因?yàn)?\angle DCB = 30^{\circ},$所以$\angle DCE = 90^{\circ},$所以$DC^{2}+EC^{2}=DE^{2},$所以$DC^{2}+BC^{2}=AC^{2},$所以四邊形$ABCD$是勾股四邊形。
(3)將$\triangle ABC$繞頂點(diǎn)$B$按逆時(shí)針?lè)较蛐D(zhuǎn)$60^{\circ},$使點(diǎn)$C$與點(diǎn)$D$重合,得到$\triangle EBD,$連接$AE,$所以$AB = BE,$$AC = DE,$$\angle ABE = 60^{\circ},$所以$\triangle ABE$是等邊三角形,所以$AE = AB,$$\angle EAB = 60^{\circ}。$因?yàn)?\angle DAB = 30^{\circ},$所以$\angle DAE = \angle DAB+\angle BAE = 30^{\circ}+60^{\circ}=90^{\circ},$所以$\triangle DAE$為直角三角形,所以$DE^{2}=AD^{2}+AE^{2},$即$AC^{2}=AD^{2}+AB^{2},$所以$AC^{2}=8^{2}+6^{2}=100,$所以$AC = 10。$