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In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC^{2} = AD^{2 }+ BC x DE + `(1)/(4)"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.

In ΔADC, ∠ADC is an obtuse angle.

∴ AC^{2} = AD^{2} + DC^{2 }+ 2 x DC x DE

⇒ AC^{2} = AD^{2} + `(1/2"BC")^2 + 2 xx (1)/(2)"BC" xx "DE"`

⇒ AC^{2} = AD^{2} + `(1)/(4)"BC"^2 + "BC" xx "DE"`

⇒ AC^{2 }= AD^{2} + BC x DE + `(1)/(4)"BC"^2` . ....(i)

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