Question

46. In the given figure, D is mid-point of AB,
DE//BC and DF//AC. Prove that : DE = BF.




F.
С

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ĄNsWeR࿐}}}$

Given A △ABC in which D is the mid-point of side BC and ED is the bisector of ∠ADb, meeting AB in E, EF is drawn parallel to BC meeting AC in F. 

To proof ∠EDF is a right angle.

Proof In △ADB, DE is the bisector of ∠ADB.

∴    DBAD=EBAE

⇒  DCAD=EBAE.......(i)         [∵ D is the mid-point of BC ∴ DB=DC]

In △ABC, we have

         EF∣∣BC

⇒   DCAD=FCAF

⇒   In △ADC, DF divides AC in the ratio AD:DC

⇒   DF is the bisector of ∠ADC

Thus, DE and DF are the bisectors of adjacent supplementary angles ∠ADB and ∠ADCrespectively.

Hence, ∠EDF is a right angle.

HOPE SO IT WILL HELP........