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The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2. Q.14

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For class 10th Triangles chapter how i solve the problem from exercise 6.5 of question no.14, give me the best way to solve the problem of this question The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.

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  1. Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

    DB = 3CD.

    In Δ ABC,

    AD ⊥BC and BD = 3CD

    In right angle triangle, ADB and ADC, by Pythagoras theorem,

    AB2 = AD2 + BD2 ……………………….(i)

    AC2 = AD2 + DC2 ……………………………..(ii)

    Subtracting equation (ii) from equation (i), we get

    AB2 – AC2 = BD2 – DC2

    = 9CD2 – CD2 [Since, BD = 3CD]

    = 8CD2

    = 8(BC/4)[Since, BC = DB + CD = 3CD + CD = 4CD]

    Therefore, AB2 – AC2 = BC2/2

    ⇒ 2(AB2 – AC2) = BC2

    ⇒ 2AB2 – 2AC2 = BC2

    ∴ 2AB2 = 2AC2 + BC2.

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