Experts pls solve the following question

Dear student,

For circumcentre we have to show AD is perpendicular bisector of BC.
In △ABD and △ADC,
AB = AC (given)
AD = AD (common)
BD = DC (D is midpoint of BC)
△ABD ≅△ADC (BY SSS congruency)
⇒ ∠ADB + ∠ADC = 180°
⇒ AD I BC,
⇒ BD = DC
So AD is perpendicular bisector of BC.
So, the circumcentre lie on AD.
For incentre we have to show AD is bisector of ∠BAC.
Since △ABD ≅ △ADC
⇒ ∠BAD = ∠CAD (By CPCT)
⇒ AD is the bisector of ∠BAC.
Hence, incenter lies on AD.
For orthocenter we need to prove AD is altitude corresponding to side BC.
Since △ABD ≅ △ADC
⇒ ∠ADB + ∠ADC = 180°
⇒ AD I BC,
⇒ AD is altitude corresponding to side BC.
For centroid we have to prove that AD is median corresponding to BC.
Since, it is given that D is the midpoint of BC.
AD is the median.
So, centroid lies on AD.
Hence Proved.

Regards.