Q20 of 95 Page 5

In ΔABC, ∠B = 90° and D is the midpoint of BC. Prove that AC2 = AD2 + 3CD2.


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


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