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

Question

Given any triangle $ˆ A B C$ $hat(ABC)$ , stretch $¯ ¯¯¯¯ ¯ A C$ $bar(AC)$ to D such that $¯ ¯¯¯¯ ¯ C D ≅ ¯ ¯¯¯¯ ¯ C B$ $bar(CD)≅bar(CB)$ ; stretch also $¯ ¯¯¯¯ ¯ C B$ $bar(CB)$ into E such that bar(CE)≅bar(CA).Segment $¯ ¯¯¯¯ ¯ D E and ¯ ¯¯¯¯ ¯ A B$ $bar(DE) and bar(AB)$ meet at F. Show that $ˆ D F B$ $hat(DFB$ is isoscheles?

2005 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-19T16:15:55

As follows

Ref:Given Figure

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

$"Again in "DeltaABC and DeltaDEC$

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

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

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

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

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

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