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In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2= AB2+ BC2 – 2 BC.BD. Q.4

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For class 10 Triangles chapter of exercise 6.6 of ncert math, how i solve this chapter because it is an optional exercise and i don’t know how to solve this problem, In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2= AB2+ BC2 – 2 BC.BD.

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  1. By applying Pythagoras Theorem in ∆ADB, we get,

    AB2 = AD2 + DB2

    We can write it as;

    ⇒ AD2 = AB2 − DB2 ……………….. (i)

    By applying Pythagoras Theorem in ∆ADC, we get,

    AD2 + DC2 = AC2

    From equation (i),

    AB2 − BD2 + DC2 = AC2

    AB2 − BD2 + (BC − BD) 2 = AC2

    AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD

    AC2 = AB2 + BC2 − 2BC × BD

    Hence, proved.

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