13. 在梯形 $ A B C D $ 中,$ A D // B C $,$ A D ＜ B C $,$ A B = 5 $,$ B C = 5 $,$ A C = 6 $.
(1) 若梯形 $ A B C D $ 是直角梯形,求 $ C D $ 的长;
(2) 设 $ A D = x $,$ C D = y $,求 $ y $ 与 $ x $ 之间的关系式,并写出 $ x $ 的取值范围;
(3) 当梯形 $ A B C D $ 是等腰梯形时,已知 $ ∠ A B C = ∠ D C B $,在直线 $ C D $ 上取一点 $ P $,使得 $ △ B C P $ 是以 $ B C $ 为腰的等腰三角形,直接写出此时 $ △ B C P $ 的底边长.
答案：13.(1)
∵AB = 5,BC = 5,AC = 6,
∴△ABC不可能是直角三角形,即AB与CB不可能垂直.
∵梯形ABCD是直角梯形,
∴BC⊥CD,如图①,过点B作BH⊥AC.
∵AB = BC = 5,
∴AH = HC = $\frac{1}{2}$AC = 3,
∴BH = $\sqrt{AB^{2}-AH^{2}}$ = 4.过点A作AE⊥BC,则$\frac{1}{2}×BC×AE = \frac{1}{2}×AC×BH$,即$\frac{1}{2}×5AE = \frac{1}{2}×6×4$,解得AE = $\frac{24}{5}$.
∵AD//BC,AE⊥BC,BC⊥CD,
∴四边形AECD是矩形,
∴CD = AE = $\frac{24}{5}$.
(2)如图②,过点A作AE⊥BC,过点D作DF⊥BC,由(1)可知AE = DF = $\frac{24}{5}$,BE = $\sqrt{AB^{2}-AE^{2}}$ = $\frac{7}{5}$,
∴EC = BC - BE = $\frac{18}{5}$.
∵AD = x,CD = y,
∴EF = x,
∴FC = $\frac{18}{5}-x$,在Rt△DFC中,DC² = DF² + FC²,
∴y² = ($\frac{24}{5}$)² + ($\frac{18}{5}-x$)²,整理得y² = x² - $\frac{36}{5}$x + 36.同理可知,当点F在点C处或BC的延长线上时,x与y同样满足上述关系式,则y² = x² - $\frac{36}{5}$x + 36(0＜x＜5).
(3)△BCP的底边长为6或$\frac{14}{5}$或8. 解析:①当点P在C,D之间时,△BCP是以BC为腰的等腰三角形,则BC = BP,如图③,过点A作AE⊥BC,过点B作BN⊥DC,由题意知∠ABC = ∠BCN,AB = CB,又
∵∠AEB = ∠BNC,
∴△ABE≌△BCN(AAS),
∴CN = BE = $\frac{7}{5}$,
∴底边PC = 2CN = $\frac{14}{5}$.
②如图④,当点D与点P重合时,BC = PC,△BCP是以BC为腰的等腰三角形,此时底边BD = AC = 6.
③如图⑤,当点P在DC的延长线上时,△BCP是以BC为腰的等腰三角形,则BC = CP,连接BD.
∵BC = CD = PC = 5,
∴∠DBC = ∠BDC,∠CBP = ∠P,
∴∠DBC + ∠CBP = 90°,即∠DBP = 90°.
∵BD = 6,DP = 10,
∴BP = $\sqrt{DP^{2}-BD^{2}}$ = 8.综上所述,△BCP底边的长为6或$\frac{14}{5}$或8.
