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Question
D is the mid point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that c2 = `"p"^2 - "a"x + "a"^2/4`
Solution
From the figure, D is the midpoint of BC.
We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse,
In ∆ABD, ∠ADE is an acute angle.
AB2 = AD2 + BD2 – 2BD . DE
⇒ AB2 = AD2 + (12BC)2 – 2 × 12 BC . DE
⇒ AB2 = AD2 + 14 BC2 – BC . DE
⇒ AB2 = AD2 – BC . DE + 14 BC2
⇒ c2 = p2 – ax + 14 a2
Hence proved.
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