Given: ∠B = 90° and D is the midpoint of BC .i.e. BD = DC
To Prove: AC2 = AD2 + 3CD2
In ∆ABC, using Pythagoras theorem we have,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AB)2 + (BC)2 = (AC)2
⇒ (AB)2 + (2CD)2 =(AC)2
⇒ (AB)2 + 4(CD)2 =(AC)2
⇒ (AD2 – BD2) + 4(CD2) = AC2
[∵ In right triangle ∆ABD, AD2 =AB2 + BD2 ]
⇒ AD2 – BD2 + 4CD2 = AC2
⇒ AD2 – CD2 + 4CD2 = AC2
[∵ D is the midpoint of BC, BD = DC]
⇒ AD2 +3CD2 = AC2
or AC2 = AD2 + 3CD2
Hence Proved