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