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Question
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2AD2 + `(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
AC2 + AB2 = AD2 + BC x DE + `(1)/(4)"BC"^2 + "AD"^2 - "BC" xx "DE" + (1)/(4)"BC"^2`
⇒ AB2 + AC2 = `2"AD"^2 + (1)/(2)"BC"^2`. ....(iii)
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