Q) If in Δ ABC, AD is median and AE ⊥ BC, then prove that AB² + AC² = 2 AD² + BC²

Ans: 

Step 1: Let’s make a diagram for the question:

Step 2: In Δ ABE, ∠ AEB is 90⁰, 

∴ by Pythagoras theorem: AB² = AE² + BE²    …….. (i) 

Similarly, In Δ ACE, ∠ AEC is 90⁰,

∴ by Pythagoras theorem: AC² = AE² + CE² ……… (ii) 

Step 3: By adding equations (i) and (ii), we get:

AB² + AC² = (AE² + BE² ) + (AE² + CE²)

∴ AB² + AC² = 2 AE² + BE² + CE² ………….(iii)

Step 4: In Δ ADE, ∠ AED is 90⁰, 

∴ by Pythagoras theorem: AD² = AE² + DE²   

∴ AE² = AD² – DE²    …….. (iv) 

Since BE = BD + DE

∴ BE² = (BD + DE)²    = BD² + DE²  + 2 BD . DE …….. (v)

Similarly, since CE = CD – DE

∴ CE² = (CD – DE)²    = CD² + DE²  – 2 CD . DE …….. (vi)

Step 5: By substituting values from equations (iv), (v) and (vi) into equation (iii), we get:

∴ AB² + AC² = 2 AE² + BE² + CE² 

= 2 (AD² – DE² ) + (BD² + DE²  + 2 BD . DE) + (CD² + DE²  – 2 CD . DE)

= 2 AD² – + BD² + + 2 BD . DE + CD² + – 2CD . DE

∴ AB² + AC² = 2 AD²  + BD² + 2 BD . DE + CD² – 2 CD . DE ……………. (vii)

Step 6: Since D is the mid-point of BC, hence, BD = CD

∴ AB² + AC² = 2 AD²  + BD² + 2 BD . DE + (BD)² – 2 (BD) . DE

= 2 AD²  + BD² + + BD² –

∴ AB² + AC² = 2 AD²  + 2 BD² …………….. (viii)

Step 7: Since D is the mid-point of BC, hence, BC = 2 BD

∴ BD =

∴ AB² + AC² = 2 AD²  + 2

∴ AB² + AC² = 2 AD²  + BC²

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