Question

In a triangle ABC , D is the midpoint of BC and ED is the bisector of the angle ADB and EF is the drawn parallel to BC cutting AC in F . Prove that angle EDF is a right angle .

Answer 1

Answer:

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.

∴

DB

AD

=

EB

AE

⇒

DC

AD

=

EB

AE

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

In △ABC, we have

EF∣∣BC

⇒

DC

AD

=

FC

AF

⇒ 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