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Question Number 13067 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17

in triangle ABC:  i)       ∡BIA=∡CIB=∡AIC  ii)    DE⊥BC,DF⊥AC,DH⊥AB  iii)   S_(ABC) =  1  (area of ΔABC)  1) ID=?  2)  ((DE)/(AI))+((DH)/(CI))+((DF)/(BI))=?  3)if S_(AHDF) =S_(CFDE) +S_(BHDE) ⇒∡HDF=?  (S_(AHDF) =area of AHDF)

intriangleABC:i)BIA=CIB=AICii)DEBC,DFAC,DHABiii)SABC=1(areaofΔABC)1)ID=?2)DEAI+DHCI+DFBI=?3)ifSAHDF=SCFDE+SBHDEHDF=?(SAHDF=areaofAHDF)

Commented by mrW1 last updated on 13/May/17

The position of I is uniquely defined,  but the position of D is not uniquely  defined, therefore ID could be any  value.    ∡HDF=180°−∡A, this is always true.

ThepositionofIisuniquelydefined,butthepositionofDisnotuniquelydefined,thereforeIDcouldbeanyvalue.HDF=180°A,thisisalwaystrue.

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17

additional conditions:  BE=EC   ,DE=1 .  special case: D ≅I  note:I ,D ,E ,are not collinear in general.

additionalconditions:BE=EC,DE=1.specialcase:DInote:I,D,E,arenotcollinearingeneral.

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/May/17

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