(1)
∵$AB \perp BD,$$AB = 3,$$CD = x,$$BD = 12,$
∴$BC = BD - CD = 12 - x。$
在$Rt\triangle ABC$中,由勾股定理得:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + (12 - x)^2} = \sqrt{9 + (12 - x)^2}。$
∵$ED \perp BD,$$DE = 2,$在$Rt\triangle DEC$中,由勾股定理得:
$CE = \sqrt{CD^2 + DE^2} = \sqrt{x^2 + 2^2} = \sqrt{x^2 + 4}。$
∴$AC + CE = \sqrt{9 + (12 - x)^2} + \sqrt{x^2 + 4}。$
(2)當(dāng)點(diǎn)$C$是線段$AE$與$BD$的交點(diǎn)時(shí),$AC + CE$的值最小。
理由:兩點(diǎn)之間線段最短,此時(shí)$AC + CE = AE。$
過(guò)點(diǎn)$A$作$AF // BD,$交$ED$的延長(zhǎng)線于點(diǎn)$F,$則四邊形$ABDF$為矩形,
∴$AF = BD = 12,$$DF = AB = 3,$$EF = DE + DF = 2 + 3 = 5。$
在$Rt\triangle AFE$中,由勾股定理得:
$AE = \sqrt{AF^2 + EF^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13。$
∴$AC + CE$的最小值為$13。$
(3)構(gòu)造幾何模型:設(shè)線段$BD = 8,$點(diǎn)$C$為$BD$上一動(dòng)點(diǎn),設(shè)$CD = x,$則$BC = 8 - x。$
過(guò)點(diǎn)$B$作$AB \perp BD,$且$AB = 5;$過(guò)點(diǎn)$D$作$ED \perp BD,$且$ED = 1,$連接$AC$、$CE。$
則$AC = \sqrt{BC^2 + AB^2} = \sqrt{(8 - x)^2 + 5^2} = \sqrt{(8 - x)^2 + 25},$
$CE = \sqrt{CD^2 + DE^2} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1},$
∴代數(shù)式$\sqrt{x^2 + 1} + \sqrt{(8 - x)^2 + 25}$的最小值即為$AC + CE$的最小值。
連接$AE,$交$BD$于點(diǎn)$C,$此時(shí)$AC + CE = AE$最小。
過(guò)點(diǎn)$A$作$AF // BD,$交$ED$的延長(zhǎng)線于點(diǎn)$F,$則四邊形$ABDF$為矩形,
∴$AF = BD = 8,$$DF = AB = 5,$$EF = DE + DF = 1 + 5 = 6。$
在$Rt\triangle AFE$中,由勾股定理得:
$AE = \sqrt{AF^2 + EF^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10。$
∴代數(shù)式$\sqrt{x^2 + 1} + \sqrt{(8 - x)^2 + 25}$的最小值為$10。$