Question

please solve this for me​

Answer 1

Step-by-step explanation:

Proof: ∆ ABD ≅ ∆ ACD

In ∆ ABD and ∆ ACD ,

AD = AD (common)

AB = AC (given)

BD = DC (given)

∴ ∆ ABD ≅ ∆ ACD (S.S.S Axiom)

Hence proved.