Question

BD and CD are the bisectors of ∠ABC and ∠ACB respectively. Find ∠BDC
PLZ HURRY!!

Answer 1

Answer:

From above given figure , we have

In ΔABC, ∠ACE = ∠ABC + ∠BAC

Similarly in ΔBCD, ∠BDC = ∠DCE − ∠DBC [Ext. angle prop. of a Δ]

But ∠DCE =1∠ACE and2

⇒1∠DBC =1∠ABC22

Now ∠BDC = ∠DCE - ∠DBC

=1(∠ACE - ∠ABC)2

=1(∠ACE + ∠ABC - ∠ACE)2∴ ∠BDC =1∠BAC2