In ΔADC ∠DAC = ∠DCA ⇒ CD= AD
Remember: Equal sides have equal angles opposite to them.
Therefore, ΔADC is an isosceles triangle.
Hence, statement 1 is correct.
In ΔABC,
AB = BC ⇒ ∠ACB = ∠BAC
∴ ∠BAC – ∠DAC = ∠ACB – ∠DCA
⇒ ∠BAD = ∠BCD
In ΔABD and ΔCBD,
AD = CD (Proved)
∠BAD = ∠BCD (Proved)
AB = BC (Given)
∴ ΔABD ≅ ΔCBD (SAS congruence rule)
Hence, statement 3 is correct.
Now, D can be any point in the interior of the ΔABC with
∠DAC = ∠DCA or CD = AD
So, BD is not always equal to AD and DC
Hence, D is not the centroid of ΔABC.
Hence, statement 2 is not correct.