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 - Mathematics

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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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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.
∴ 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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Chapter 17: Pythagoras Theorem - Exercise 17.1

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Frank Class 9 Maths ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 15.1

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