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In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB^{2} + AC^{2} = 2AD^{2 }+ `(1)/(2)"BC"^2`

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#### Solution

We have ∠AED = 90°

∴ ∠ADE < 90° and ∠ADC > 90°

i.e. ∠ADE is acute and ∠ADC is obtuse.

Adding (i) and (ii), we have

AC^{2} + AB^{2 }= AD^{2} + BC x DE + `(1)/(4)"BC"^2 + "AD"^2 - "BC" xx "DE" + (1)/(4)"BC"^2`

⇒ AB^{2} + AC^{2} = `2"AD"^2 + (1)/(2)"BC"^2`. ....(iii)

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