<prompt> Katika pembetatuABC, angle ABC120circ,AB3 na BC4. Ikiwa vipengele vilivyoundwa hadi overlineAB kwa A na kwa overlineBC kwa C vinakutana kwa D, basi CDtextA 3quad textB frac8sqrt3quad textC 5quad textD frac112quad textE frac10sqrt3  <response>: Tunaanza kwa kuchora mchoro.asy import olimpiad; kuagiza cse5; kuagiza jiometri; ukubwa 150; defaultpenfontsize10pt; defaultpen 0.8; dotfactor 4; jozi A asili; jozi C Adir55; jozi D Adir0; jozi B uganiA,Adir90,C,Cdir-155; leboA,A,S; leboC,C,NE; leboD,D,SE; leboB,B,NW; lebo4,B--C,NW; lebo3,A--B,W; kuchoraA--C--D--mzunguko; choraA--B--C; choraalama ya kuliaB,C,D,2; choraalama ya kuliaB,A,D,2; asyTunaongeza CB na DA kukutana kwa E. Hii inatupa pembetatu mbili za kulia katika CED na BEA.asy kuleta olympiad; kuagiza cse5; kuagiza jiometri; ukubwa 250; defaultpenfontsize10pt; defaultpen 0.8; dotfactor 4; jozi A asili; jozi C Adir55; jozi D Adir0; jozi B uganiA,Adir90,C,Cdir-155; jozi E uganiA,A2dir180,B,B2dir-155; leboA,A,S; leboC,C,NE; leboD,D,SE; leboB,B,NW; lebo4,B--C,NW; lebo3,A--B,W; leboE,E,SW; kuchoraA--C--D--mzunguko; choraA--B--C; choraalama ya kuliaB,C,D,2; choraalama ya kuliaB,A,D,2; choraA--E--B,iliyopigwa; asyTunaona kwamba angle E 30circ. Kwa hivyo, pembetatu BEA na pembetatu DEC ni pembetatu 30-60-90.Kwa kutumia uwiano wa kando wa pembetatu 30-60-90, tuna BE2BA6. Hii inatuambia kuwa CEBCBE4610. Pia, EA3sqrt3.Kwa sababu triangle DECsimtriangle BEA, tunafrac103sqrt3fracCD3.Kutatua mlingano, tumeanza align fracCD3frac103sqrt3 CD3cdotfrac103sqrt3 CDfrac 10sqrt3 endalign 