Q. 83.9( 8 Votes )

# In a right ∆ABC,

Answer :

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

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