Answer:
You have to prove that the triangles ABD and CBD are congruent.
Here's how you do it:
[tex]\large\text{1. AB = CB (S) [Given]$}[/tex]
[tex]\large\text{2. AD = CD (S) [D is midpoint of AC.]$}[/tex]
[tex]\large\text{3. BD = BD (S) [Common side in both triangles ABD and CBD.]$}[/tex]
[tex]\large\text{4. $\triangle$ABD $\cong\triangle$CBD [By S.S.S. axiom.]}[/tex]
Now to prove AC ⊥ BD,
[tex]\large\text{5. $\angle$ADB = $\angle$CDB [Corresponding angles of congruent triangles}[/tex]
[tex]\large\text{are equal.]$}[/tex]
[tex]\large\text{6. $\angle$ADB + $\angle$CDB$ =180^\circ$ [Sum of angles of straight line is $180^\circ.$]}[/tex]
[tex]\large\text{7. $\angle$CDB + $\angle$CDB = $180^\circ$ [From statement 5.]}[/tex]
[tex]\large\text{$\angle$CDB = $90^\circ$}[/tex]
[tex]\large\text{i.e. AC $\perp$ BD}[/tex]