Let #hat(ABC)# be any triangle, stretch #bar(AC)# to D such that #bar(CD)≅bar(CB)#; stretch also #bar(CB)# into E such that bar(CE)≅bar(CA). Segments #bar(DE) and bar(AB)# meet at F. Show that #hat(DFB# is isosceles?

Given any triangle #hat(ABC)#, stretch #bar(AC)# to D such that #bar(CD)≅bar(CB)#; stretch also #bar(CB)# into E such that bar(CE)≅bar(CA).Segment #bar(DE) and bar(AB)# meet at F. Show that #hat(DFB# is isoscheles?
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1 Answer
Jun 19, 2016

As follows

Explanation:

Ref:Given Figure

#"In " DeltaCBD,bar(CD)~=bar(CB)=>/_CBD=/_CDB#

#"Again in "DeltaABC and DeltaDEC#

#bar(CE)~=bar(AC)->"by construction"#

#bar(CD)~=bar(CB)->"by construction"#

#"And "/_DCE=" vertically opposite "/_BCA#

#"Hence " DeltaABC~=DeltaDCE#
#=>/_EDC=/_ABC#

#"Now in "DeltaBDF,/_FBD=/_ABC+/_CBD=/_EDC+/_CDB=/_EDB=/_FDB#

#"So "bar(FB)~=bar(FD)=>DeltaFBD " is isosceles"#