Let's see how many of u can.....!

Shouldnt it be BD²=AD.CD ?

If BD is the median to AC.
AD=DC
Now I have drawn ED||BC.
Then by mid-point theorem, E is the mid-point of AB.
So, AE=EB.
In Triangle ADE and Triangle ABC
AE/AB=AD/AC=1/2
So, Triangle ADE is SIMILAR toTriangle ABC.

Therefore, Angle AED= Angle ABC=90°

In triangle ADE and triangle EDB.
ED is common.
AE=EB (Proved)
Angle AED=Angle DEB=90°(Proved)
So, Triangle ADE is CONGRUENT to Triangle DEB.

So, AD=DB (C.P.C.T.)

So, AD=BD=DC

@everyone -> Guys i have missed a very important part in my question, BD is the median on AC i.e... AD = CD

@Arnab -> Why have you used similarity.... i mean you are absolutely right...

But what was the use of similarity...????

Simple :

When you draw DE || BC then LAED = LABC = 90⁰.... (corr. Ls)

and also by Thales' theorm or mid-point theorem...

AE = BE

and so triangles... AED and BED are congruent (by SAS) and so...

BD = AD and so...

AD = BD = CD (simple)

simpler solution

AD=CD

D is the circumcenter

result follows