Question

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?

Answer 1

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"`