In a right ∆ABC,

It is given in the question that,

In right triangle ABC, ∠B = 90^o

Also D is the mid-point of AC

∴ AD = DC

∠ADB = ∠BDC (BD is the altitude)

BD = BD (Common)

So, by SAS congruence criterion

∴ ∆ADB ≅ ∆CDB

∠A = ∠C (CPCT)

As, ∠B = 90^o

So, by using angle sum property

∠A = ∠ABD = 45^o

Similarly, ∠BDC = 90^o (BD is the altitude)

∠C = 45^o

∠DBC = 45^o

∠ABD = 45^o

Now, by isosceles triangle property we have:

BD = CD and

BD = AD

AS, AD + DC = AC

BD + BD = AC

2BD = AC

BD =

Hence, proved