Sum

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 c^{2} = `"p"^2 - "a"x + "a"^2/4`

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#### 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.

AB^{2} = AD^{2} + BD^{2} – 2BD . DE

⇒ AB^{2} = AD^{2} + (12BC)2 – 2 × 12 BC . DE

⇒ AB^{2} = AD^{2} + 14 BC^{2} – BC . DE

⇒ AB^{2} = AD^{2} – BC . DE + 14 BC^{2}

⇒ c^{2} = p^{2} – ax + 14 a^{2}

Hence proved.

Concept: Similarity

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