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In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.

Prove that: 2AB^{2 }= 2AC^{2 }+ BC^{2}

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#### Solution

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In ∆ABC, AD ⊥ BC, and BD = 3CD ...(Given)

In ∆ADC, ∠ADC = 90°

By Pythagoras' theorem,

AC^{2} = AD^{2} + CD^{2}

2AC^{2} = 2AD^{2} - 2CD^{2} ...(1) [Multiplied by 2]

In ∆ADB, ∠ADB = 90°

by Pythagoras' theorem,

AB^{2} = AD^{2} + BD^{2}

2AB^{2} = 2AD^{2} + 2BD^{2} ...(2) [Multiplied by 2]

Subtracting equation (1) from (2)

2AB^{2} - AC^{2} = (2AD^{2} + 2BD^{2}) - (2AD^{2} + 2CD^{2})

= 2AD^{2} + 2BD^{2} - 2AD^{2} - 2CD^{2}

= 2(3CD)^{2} - 2CD^{2 }

= 2 × 9CD^{2} - 2CD^{2} ...[Given]

= 18CD^{2} - 2CD^{2}

= 16CD^{2}

∴ 2AB^{2} - 2AC^{2} = 16CD^{2} ...(3)

BC = CD + DB [C-D-B]

BC = CD + 3CD = 4CD

BC^{2} = 16CD^{2} ...(4) [squaring both sides]

From (3) & (4)

2AB^{2} - 2AC^{2} = BC^{2}

∴ 2AB^{2} = 2AC^{2} + BC^{2}

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