Question

In triangle ABC , D is the mid point of BC and ED is the bisector of angle ADB and EF is drawn parallel to BC cutting AC at F.prove that the angle EDF is a right angle.

Answer 1

BD = DC

∡ADE = ∡EDB

FE ║ CB  ⇒ ∡BDE = ∡FED

∡FED = ∡ADE = α  ⇒ MD = ME,  where M is the crosspoint of drawns AD and FE

FE ║ CB  ⇒  CD/DB = FM/ME = 1 ⇒ FM = ME (M is the midpoint of FE)

FM = MD  ⇒ ∡MFD = ∡MDF = β

∡FED + ∡EFD = α + β

∡ADE + ∡ADF = α + β

α + β + α + β = 2(α+ β) = 180°

α + β = 90° = ∡EDF

So ∡EDF is a right angle.