Question

Please solve this question​

Answer 1

Step-by-step explanation:

in ∆ ABC and OUTER side CE

<ACE = < CAB +< ABC----1

in ∆ BDC and outer side CE

<DCE =<CBD +<BDC ----- 2

NOW FROM 1

2<DCE = < BAC + 2< CBD [ as ,< ABD = <CBD and < ACD = < DCE]

i.e 2 ( < CBD + < BDC ) = < BAC + 2 < CBD

(put the value of equation 2)

i.c 2 < CBD + 2< BDC = < BAC + 2 < CBD

i.c 2 < BDC = < BAC

i.e < BDC = 1/2 <BAC