Question

The angle ACD is the exterior angle of Δ ABC and the bisector of angle A intersects BC in E. Prove that angle ABC + angle ACD = 2 (angle AEC)

Answer 1

Step-by-step explanation:

HOPE THIS ATTACHMENT HELPS..

Here angle ABC is written as angle B for convenience....don't get confused both are the same angles

Answer 2

Step-by-step explanation:

Given ∆ABC, AE is the angle bisector ( median )

Exterior Angle is the sum of opposite interior angles.

So, Angle ACD = Angle ABC + Angle BAC

Angle BAC = Angle BAE + Angle EAC

[Angle BAE = Angle EAC, because AE is angle bisector]

=> Angle BAC = 2(Angle BAE)

So, Angle ACD = Angle ABC + 2(Angle BAE)

To proof : Angle ABC + Angle ACD = 2 (Angle AEC)

LHS

Angle ABC + Angle ACD

=> Angle ABC + Angle ABC + 2(Angle BAE)

= 2 ( Angle ABC ) + 2 ( Angle BAE )

= 2 ( Angle ABC + Angle BAE )

Angle AEC is also exterior angle of ∆BAE

Therefore,

Angle AEC = Angle ABC + Angle BAE.......

Angle ABC + Angle ACD = 2 ( Angle ABC + Angle BAE )

Substituting Angle AEC = Angle ABC + Angle BAE , we get,

Angle ABC + Angle ACD = 2 ( Angle AEC )

Hence Proved