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प्रश्न
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`
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उत्तर
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.
∴ AC2 = AD2 + DC2 + 2 x DC x DE
⇒ AC2 = AD2 + `(1/2"BC")^2 + 2 xx (1)/(2)"BC" xx "DE"`
⇒ AC2 = AD2 + `(1)/(4)"BC"^2 + "BC" xx "DE"`
⇒ AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2` . ....(i)
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