Question

I need it urgently..​

Answer 1

Step-by-step explanation:

In ∆ ABC,

AB = AC,

therefore, < ABC = < ACB ______ EQUATION 1

and,

in ∆ DBC,

DB = DC

therefore, < DBC = < DCB ______ EQUATION 2

Adding EQUATION 1 and EQUATION 2,

< ABC + < DBC = < ACB + < DCB

Now, < ABC + < DBC = < ABD

and, < ACB + < DCB = < ACD

so, from above,

< ABD = < ACD

Hence proved....

Answer 2

Answer:

triangle abc is isosceles i.e. ab=ac

thus angle abc =angle acb (angles opp. to equal side are equal)....(1)

also triangle dbc is isosceles i.e. db=dc

thus angle dbc=angle dcb (angles opp. to equal side are equal).....(2)

hence adding (1) and (2) we get angle abd=angle acd

hence proved