Question

In the given figure AD is internal bisector of angle A and CE is parallel to DA. If CE meets BA produced at E prove that triangle CAE is isosceles. ​

Answer 1

Answer:

Δ CAE is an isosceles triangle

Step-by-step explanation:

In the given figure AD is internal bisector of angle A and CE is parallel to DA. If CE meets BA produced at E prove that triangle CAE is isosceles

∠BAC  + ∠EAC = 180° ( Straight line)

in ΔACE

∠ACE  + ∠AEC + ∠EAC = 180°  ( sum of angles of Triangle)

Equating both

∠BAC  + ∠EAC = ∠ACE  + ∠AEC + ∠EAC

=> ∠BAC  = ∠ACE  + ∠AEC    Eq 1

∠BAD = (1/2) ∠BAC  

∠BAD = ∠AEC  ( AD ║ CE)

=> ∠AEC = (1/2) ∠BAC

putting this in eq 1

=> ∠BAC  = ∠ACE  + (1/2) ∠BAC

=> ∠ACE = (1/2) ∠BAC

∠AEC = ∠ACE = (1/2) ∠BAC

=> AC = AE  

Hence Δ CAE is an isosceles triangle

Answer 2

Answer:

CAE is an isosceles triangle