Q. 274.8( 5 Votes )

# In ΔABC, D is the mid-point of BC and AE ⊥ BC . If AC > AB, show that AB^{2} = AD^{2} — BC .DE + 1/4 BC^{2}

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} = a^{2} + b^{2} – 2ab]

⇒ (AE^{2} + ED^{2}) + (BD)^{2} – 2 (ED)(BD) = (AB)^{2}

⇒ (AD)^{2} + (BD)^{2} – 2 (ED)(BD) = (AB)^{2}

[∵ In right angled ∆AED, AE^{2} + ED^{2} =AD^{2}]

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

Hence Proved

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