Q. 274.8( 5 Votes )

# In ΔABC, D is the mid-point of BC and AE ⊥ BC . If AC > AB, show that AB2 = AD2 — BC .DE + 1/4 BC2

Answer :

Given: In ABC, D is the mid-point of BC and AE BC

and AC > AB

In right triangle ∆AEB, using Pythagoras theorem, we have

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AE)2 + (BE)2 = (AB)2

(AE)2 + (BD – ED)2 = (AB)2

(AE)2 + (ED)2 + (BD)2 – 2 (ED)(BD) = (AB)2

[ (a – b)2 = a2 + b2 – 2ab]

(AE2 + ED2) + (BD)2 – 2 (ED)(BD) = (AB)2

(AD)2 + (BD)2 – 2 (ED)(BD) = (AB)2

[ In right angled ∆AED, AE2 + ED2 =AD2]

[ D is the midpoint of BC, so 2DC = BC]

Hence Proved

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