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9. 如图,在$△ ABC$中,$∠ B=30°$,$∠ C=90°$,等边三角形$DEF$的三个顶点分别落在$AC$,$AB$,$BC$上.若$CD=4$,$BE=6$,则$AB$的长为
$\dfrac{32}{3}$
.答案
9. $\dfrac{32}{3}$
三、解答题(共42分)
10. (18分)如图,在四边形ABCD中,过点C作$CE ⊥ AB$于点E,且$CD=CB$,$∠ CBE + ∠ ADC = 180°$.
(1)若$AB=5$,$BE=1$,求AD的长;
(2)若$△ ABC$和$△ ACD$的面积分别为28和16,求$△ BCE$的面积.
10. (18分)如图,在四边形ABCD中,过点C作$CE ⊥ AB$于点E,且$CD=CB$,$∠ CBE + ∠ ADC = 180°$.
(1)若$AB=5$,$BE=1$,求AD的长;
(2)若$△ ABC$和$△ ACD$的面积分别为28和16,求$△ BCE$的面积.
答案
10. (1) 过点 $C$ 作 $CH ⊥ AD$,交 $AD$ 的延长线于点 $H$. $\because CE ⊥ AB, \therefore ∠ BEC = ∠ DHC = 90°$.
$\because ∠ CBE + ∠ ADC = 180°, ∠ CDH + ∠ ADC = 180°$,
$\therefore ∠ CBE = ∠ CDH$. 又 $\because CB = CD, \therefore △ BCE ≌ △ DCH(\mathrm{AAS}). \therefore BE = DH = 1, CE = CH$. 在 $\mathrm{Rt}△ AHC$ 和 $\mathrm{Rt}△ AEC$ 中, $\begin{cases} AC = AC, \\ CH = CE, \end{cases} \therefore \mathrm{Rt}△ AHC ≌ \mathrm{Rt}△ AEC(\mathrm{HL})$.
$\therefore AH = AE. \because AB = 5, BE = 1, \therefore AE = AB - BE = 5 - 1 = 4. \therefore AH = 4. \therefore AD = AH - DH = 4 - 1 = 3$
(2) 由(1)知, $△ BCE ≌ △ DCH, \mathrm{Rt}△ AHC ≌ \mathrm{Rt}△ AEC, \therefore S_{△ BCE} = S_{△ DCH}, S_{△ AHC} = S_{△ AEC}. \therefore S_{△ ABC} + S_{△ ACD} = S_{△ AEC} + S_{△ AHC} = 2S_{△ AEC}. \because △ ABC$ 和 $△ ACD$ 的面积分别为 28 和 16, $\therefore 2S_{△ AEC} = 28 + 16 = 44. \therefore S_{△ AEC} = 22. \therefore S_{△ BCE} = S_{△ ABC} - S_{△ AEC} = 28 - 22 = 6$
$\because ∠ CBE + ∠ ADC = 180°, ∠ CDH + ∠ ADC = 180°$,
$\therefore ∠ CBE = ∠ CDH$. 又 $\because CB = CD, \therefore △ BCE ≌ △ DCH(\mathrm{AAS}). \therefore BE = DH = 1, CE = CH$. 在 $\mathrm{Rt}△ AHC$ 和 $\mathrm{Rt}△ AEC$ 中, $\begin{cases} AC = AC, \\ CH = CE, \end{cases} \therefore \mathrm{Rt}△ AHC ≌ \mathrm{Rt}△ AEC(\mathrm{HL})$.
$\therefore AH = AE. \because AB = 5, BE = 1, \therefore AE = AB - BE = 5 - 1 = 4. \therefore AH = 4. \therefore AD = AH - DH = 4 - 1 = 3$
(2) 由(1)知, $△ BCE ≌ △ DCH, \mathrm{Rt}△ AHC ≌ \mathrm{Rt}△ AEC, \therefore S_{△ BCE} = S_{△ DCH}, S_{△ AHC} = S_{△ AEC}. \therefore S_{△ ABC} + S_{△ ACD} = S_{△ AEC} + S_{△ AHC} = 2S_{△ AEC}. \because △ ABC$ 和 $△ ACD$ 的面积分别为 28 和 16, $\therefore 2S_{△ AEC} = 28 + 16 = 44. \therefore S_{△ AEC} = 22. \therefore S_{△ BCE} = S_{△ ABC} - S_{△ AEC} = 28 - 22 = 6$
11. (24 分) 如图, $O$ 是等边三角形 $ABC$ 内一点, $∠ AOB=110°$, $∠ BOC=α$. 以 $OC$ 为边向右作等边三角形 $OCD$, 连接 $AD$.
(1) 当 $α=150°$ 时, 试判断 $△ AOD$ 的形状, 并说明理由;
(2) 当 $α$ 为多少时, $△ AOD$ 是等腰三角形?
(1) 当 $α=150°$ 时, 试判断 $△ AOD$ 的形状, 并说明理由;
(2) 当 $α$ 为多少时, $△ AOD$ 是等腰三角形?
答案
11. (1) 当 $α = 150°$ 时, $△ AOD$ 是直角三角形 理由:$\because △ OCD$ 是等边三角形, $\therefore OC = DC, ∠ ODC = ∠ OCD = 60°. \because △ ABC$ 是等边三角形, $\therefore BC = AC, ∠ ACB = 60°. \therefore ∠ ACB = ∠ OCD. \therefore ∠ ACB - ∠ OCA = ∠ OCD - ∠ OCA$, 即 $∠ BCO = ∠ ACD. \therefore △ BOC ≌ △ ADC. \therefore ∠ BOC = ∠ ADC. \because ∠ BOC = α = 150°, \therefore ∠ ADC = 150°. \therefore ∠ ADO = ∠ ADC - ∠ ODC = 150° - 60° = 90°. \therefore △ ADO$ 是直角三角形.
(2) $\because △ OCD$ 是等边三角形, $\therefore ∠ ODC = ∠ DOC = 60°$. 又 $\because ∠ AOB = 110°, ∠ BOC = α, \therefore ∠ AOD = 360° - 110° - α - 60° = 190° - α$. 由(1), 得 $∠ ADC = ∠ BOC = α, \therefore ∠ ADO = α - 60°. \because △ AOD$ 的内角和为 $180°, \therefore ∠ OAD = 180° - (190° - α) - (α - 60°) = 50°$. ① 要使 $AO = AD$, 只需 $∠ AOD = ∠ ADO$, 即 $190° - α = α - 60°, \therefore α = 125°$; ② 要使 $OA = OD$, 只需 $∠ OAD = ∠ ADO$, 即 $50° = α - 60°, \therefore α = 110°$; ③ 要使 $AD = OD$, 只需 $∠ OAD = ∠ AOD$, 即 $50° = 190° - α, \therefore α = 140°$. 综上所述, 当 $α$ 为 $110°, 125°, 140°$ 时, $△ AOD$ 是等腰三角形
(2) $\because △ OCD$ 是等边三角形, $\therefore ∠ ODC = ∠ DOC = 60°$. 又 $\because ∠ AOB = 110°, ∠ BOC = α, \therefore ∠ AOD = 360° - 110° - α - 60° = 190° - α$. 由(1), 得 $∠ ADC = ∠ BOC = α, \therefore ∠ ADO = α - 60°. \because △ AOD$ 的内角和为 $180°, \therefore ∠ OAD = 180° - (190° - α) - (α - 60°) = 50°$. ① 要使 $AO = AD$, 只需 $∠ AOD = ∠ ADO$, 即 $190° - α = α - 60°, \therefore α = 125°$; ② 要使 $OA = OD$, 只需 $∠ OAD = ∠ ADO$, 即 $50° = α - 60°, \therefore α = 110°$; ③ 要使 $AD = OD$, 只需 $∠ OAD = ∠ AOD$, 即 $50° = 190° - α, \therefore α = 140°$. 综上所述, 当 $α$ 为 $110°, 125°, 140°$ 时, $△ AOD$ 是等腰三角形
