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Question
In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.
Prove that: 2AB2 = 2AC2 + BC2
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Solution
In ∆ABC, AD ⊥ BC, and BD = 3CD ...(Given)
In ∆ADC, ∠ADC = 90°
By Pythagoras’ theorem,
AC2 = AD2 + CD2
2AC2 = 2AD2 − 2CD2 ...(1) [Multiplied by 2]
In ∆ADB, ∠ADB = 90°
by Pythagoras’ theorem,
AB2 = AD2 + BD2
2AB2 = 2AD2 + 2BD2 ...(2) [Multiplied by 2]
Subtracting equation (1) from (2)
2AB2 − AC2 = (2AD2 + 2BD2) − (2AD2 + 2CD2)
= 2AD2 + 2BD2 − 2AD2 − 2CD2
= 2(3CD)2 − 2CD2
= 2 × 9CD2 − 2CD2 ...[Given]
= 18CD2 − 2CD2
= 16CD2
∴ 2AB2 − 2AC2 = 16CD2 ...(3)
BC = CD + DB [C−D−B]
BC = CD + 3CD = 4CD
BC2 = 16 CD2 ...(4) [squaring both sides]
From (3) & (4)
2AB2 − 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2
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