Question

In triangle ABC the bisectors of B and C meet at O. EF is drawn parallel to BC through O. Prove that EF=BE+CF​

Answer 1

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.

∴ AD/DB = AE/EB

⇒ AD/DC = AE/EB .......(i) [∵ D is the mid-point of BC ∴ DB=DC]

In △ABC, we have

EF∣∣BC

⇒ AD/DC = AF/FC

⇒ 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.