《2025年同步练习上海科学技术出版社九年级数学上册沪科版》

20. 在$\triangle ABC$中,$\angle ACB = 90^{\circ}$,点$E是斜边AB$的中点,$AB = 10$,$BC = 8$,点$P在CE$的延长线上,过点$P作PQ \perp CB$,交$CB的延长线于点Q$,设$EP = x$.
(1) 如图(1),求证：$\triangle ABC \backsim \triangle PCQ$；
(2) 如图(2),连接$PB$,当$PB平分\angle CPQ$时,试求$x$的值；
(3) 如图(3),过点$B作BF \perp AB交PQ于点F$,若$\angle BEF = \angle A$,试求$x$的值.

答案：
(1)
∵∠ACB = 90°，E是AB中点，
∴CE = BE．
∴∠ABC = ∠PCQ．
∵PQ ⊥ CB，
∴∠PQC = 90° = ∠ACB．
∴△ABC ∽ △PCQ
(2)作BH ⊥ PE于点H，
∵PB平分∠CPQ，BH ⊥ PE，BQ ⊥ PQ，
∴BH = BQ．
∵△ABC ∽ △PCQ，
∴$\frac{PC}{CQ}$ = $\frac{AB}{BC}$ = $\frac{10}{8}$ = $\frac{5}{4}$．
∴CQ = $\frac{4}{5}$PC = $\frac{4}{5}$(CE + EP) = $\frac{4}{5}$(5 + x)．
∴BQ = CQ - BC = $\frac{4}{5}$x - 4．
∴BH = BQ = $\frac{4}{5}$x - 4．又易证△ABC ∽ △BCH，则可得BH = $\frac{24}{5}$．
∴$\frac{4}{5}$x - 4 = $\frac{24}{5}$．
∴x = 11
(3)
∵BF ⊥ AB，
∴∠EBF = 90° = ∠ACB．又
∵∠BEF = ∠A，
∴△EBF ∽ △ACB．
∴$\frac{BF}{BC}$ = $\frac{BE}{AC}$．
∵∠ABC + ∠FBQ = 90° = ∠ABC + ∠A，
∴∠FBQ = ∠A．又
∵∠ACB = 90° = ∠BQF，
∴△ACB ∽ △BQF．
∴$\frac{AB}{BF}$ = $\frac{AC}{BQ}$．
∴$\frac{AB}{BC}$ = $\frac{BE}{BQ}$．
∵AB = 10，BC = 8，BE = 5，
∴BQ = 4．由
(2)知$\frac{PC}{CQ}$ = $\frac{AB}{BC}$，
∵CQ = BC + BQ = 12，
∴PC = 15．
∴EP = PC - CE = 10，即x = 10

21. 如图(1),设$D为锐角三角形ABC$内一点,$\angle ADB = \angle ACB + 90^{\circ}$.
(1) 求证：$\angle CAD + \angle CBD = 90^{\circ}$；
(2) 如图(2),过点$B作BE \perp BD$,$BE = BD$,连接$EC$,若$AC \cdot BD = AD \cdot BC$,
① 求证：$\triangle ACD \backsim \triangle BCE$；
② 求$\frac{AB \cdot CD}{AC \cdot BD}$的值.

答案：
(1)延长CD交AB于点H，
∵∠ADH = ∠CAD + ∠ACD，∠BDH = ∠BCD + ∠CBD，
∴∠CAD + ∠CBD = ∠ADH + ∠BDH - ∠ACD - ∠BCD = ∠ADB - ∠ACB = 90°
(2)①
∵AC·BD = AD·BC，BD = BE，
∴$\frac{AC}{BC}$ = $\frac{AD}{BE}$．
∵∠CAD + ∠CBD = 90°，∠CBD + ∠CBE = 90°，
∴∠CBE = ∠CAD．
∴△ACD ∽ △BCE ② 由①知$\frac{AC}{BC}$ = $\frac{CD}{EC}$，∠ACD = ∠BCE，
∴∠ACB = ∠DCE．连接DE，由$\frac{AC}{CD}$ = $\frac{BC}{EC}$可得△ACB ∽ △DCE．
∴$\frac{AC}{CD}$ = $\frac{AB}{DE}$．
∵BE ⊥ BD，BE = BD，
∴DE = $\sqrt{2}$BD．
∴$\frac{AB·CD}{AC·BD}$ = $\frac{AB}{BD}$·$\frac{CD}{AC}$ = $\frac{AB}{BD}$·$\frac{DE}{AB}$ = $\frac{DE}{BD}$ = $\sqrt{2}$