Question

Bisector of angle BAC of triangle ABC and the straight line through the mid point D of the side AC and parallel to the side AB meet at a point E outside BC. Prove that angle AEC is equal to 1 right angle. ( answer it correctly with a proper explanation, spam will be reported ).​

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.