Question

In the given figure D is the mid point of AB DE/ BC and DF//AC proce that DE= BF

Answer 1

Answer:

Hey buddy here is your answer:-

Step-by-step explanation:

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.

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Answer 2

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