题目 #21095

小问详解过点作1:OON ⊥ PB于点，N (1) Q四边形ABCD是矩形,  =  = 90, = CD, A\(C \frac{AB}{RtQCD}\)在\(V \frac{RtABP}{CQ}\)和 V 中, AP =  =  = 90 A C  = CD\(\frac{AB}{V}\)  QCD(SAS),\(\frac{ABP}{BP}\frac{V}{DQ}\) =. (2)设AP =, = 5 + a．\(\frac{aAD}{PBQD}\)当四边形 是菱形时， PB = = 5，\(\frac{PD}{RtABP}\)在 V 中， 根据勾股定理得\(AP^{2}\)+\(AB^{2}\)=\(PB^{2}\)， 即\(a^{2}\)+ =\(5^{2}\)，\(\frac{4^{2}}{a}\)可得： =，\(\frac{3}{3}\)所以 D= + 5 = 8． A 2026-06-22  与e 相切于点M，\(\frac{PA}{OM} \frac{O}{PA}\) ⊥, APO = ,\(\frac{Q}{Q} \frac{BPO}{APB}\)是 的平分线，\(\frac{PO}{ON}\) =,\(\frac{OM}{e}\)Q OM为 的半径,\(\frac{O}{O}\) 为e 的半径,\(\frac{ON}{ON}\)Q ⊥ PB,PB是 e O的切线; 小问详解2 : Q CD ⊥ OM，OM为半径， CE = DE =\(\frac{1}{2}\)CD, Q PM =\(\frac{5}{4}\)CD,\(\frac{CD}{PM} = \frac{4}{5}\)\(\frac{CE}{PM} = 2\), 5 Q  = 90, = 90,\(\frac{OMP}{PM} \frac{OEC}{OMP}\)CD∥,V ∽VOEC, \(\frac{CE}{PM} = \frac{OC}{OP}\)Q PC = 6,\(\frac{2}{5} = \frac{OC}{OC}, + 6\) = 4,OC = OM = 4,\(\frac{OC}{R}\)在 tVM P中，\(\frac{O}{\sqrt{OP^{2}−OM^{2}}}\)PM = = （\(\sqrt{6+4）^{2}−4^{2}} = 2 \sqrt{21}\)， Q  = ,MFP = EFD,\(\frac{FMP}{MFP} \frac{FED}{EFD}\)MP V ∽V,\(\frac{MF}{EF} =\)E D 设MF = x， 则EF = 4 − x −\(\frac{8}{5} = \frac{12}{5}\)− x，  x = 221 ,\(\frac{12}{5}\)− x\(4 \sqrt{21} 5\)解得x =\(\frac{12}{7}\)，M = 12. F 7 2026-06-22