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Question
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)
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Solution
We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse.
From (iii), we have
AB2 + AC2 = `2"AD"^2 + (1)/(2)"BC"^2`
⇒ AB2 + AC2 = `2"AD"^2 + (1)/(2)(2 xx "CD")^2`
⇒ AB2 + AC2 = `2"AD"^2 + (1)/(2) xx 4"CD"^2`
⇒ AB2 + AC2 = 2AD2 + 2CD2
⇒ AB2 + AC2 = 2(AD2 + CD2).
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