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Question Number 174763 by dragan91 last updated on 10/Aug/22
Answered by a.lgnaoui last updated on 11/Aug/22
∠ADFsinADF=sinFAD=sinD24(1)∠EBFsinBx=sinF3=sinE8(2)∠CDEsinD3=sinCDE=sinECD(3)(2)and(3)sinE8=sinF3sinE=8sinF3sinECD=sinD3sinE=CDsinD38sinF3=CDsinD3CD=8sinFsinD∠ADFsinFAD=sinD24⇒sinFsinD=AD24CD=8AD24,CD=AD3⇒AC=4CDCDCI=CEBC=12=DEBI(DE∥BI)GD∣∣BCFEFD=FBFG⇒xFD=812=23H∈EF(BH∣∣AC)∡DCB=∡CBH∡C+∡D=∡B+∡FCE=BE⇒DE=EH=3∡HBF=∡ACD=BHEF=2×3=6⇒x=6
Commented by Tawa11 last updated on 11/Aug/22
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