Q13 of 45 Page 43

In ∆ ABC, seg AD seg BC DB = 3CD. Prove that : 2AB2 = 2AC2 + BC2


Given,


DB = 3CD


Also,


BC = CD + DB = CD + 3CD


⇒ BC = 4CD [1]


As, AD ⊥ BC, By Pythagoras theorem i.e.


(Hypotenuse)2 = (base)2 + (Perpendicular)2


In Δ ACD


AC2 = AD2 + CD2 [2]


In ΔABD


AB2 = AD2 + DB2


⇒ AB2 = AD2 + (3CD)2


⇒ AB2 = AD2 + 9CD2 [3]


Subtracting [2] from [3]


⇒ AB2 - AC2 = 9CD2 - CD2


⇒ AB2 = AC2 + 8CD2


⇒ 2AB2 = 2AC2 + 16CD2


⇒ 2AB2 = 2AC2 + (4CD)2


⇒ 2AB2 = 2AC2 + BC2 [From 1]


Hence Proved.


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