In a right ∆ABC, ∠B=90° and D is the mid-point of AC. Prove that BD=AC.
It is given in the question that,
In right triangle ABC, ∠B = 90o
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 = 90o
So, by using angle sum property
∠A = ∠ABD = 45o
Similarly, ∠BDC = 90o (BD is the altitude)
∠C = 45o
∠DBC = 45o
∠ABD = 45o
Now, by isosceles triangle property we have:
BD = CD and
BD = AD
AS, AD + DC = AC
BD + BD = AC
2BD = AC
BD =
Hence, proved