Question 14 - Pythagoras Theorem and it's important questions - Chapter 6 Class 10 Triangles

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Question 14 The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see figure). Prove that 2AB2 = 2AC2 + BC2 Given: ΔABC with AD ⊥ BC Also DB = 3 CD To prove: 2AB2 = 2AC2 + BC2 Proof: Let BC = x Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Now in right ∆ 𝐴𝐷𝐶 AC2 = AD2 + DC2 AC2 = AD2 + (𝑥/4)^2 AC2 = AD2 + 𝑥2/16 Similarly in right ∆ 𝐴𝐷𝐵 AB2 = AD2 + DB2 AB2 = AD2 + ( 3𝑥/4 )2 AB2 = AD2 + 9𝑥2/16 Subtracting (2) from (1) AC2 – AB2 = AD2 + 𝑥2/16 – (𝐴𝐷2+9𝑥2/16) AC2 – AB2 = AD2 + 𝑥2/16−𝐴𝐷2−9𝑥2/16 AC2 – AB2 = AD2 – AD2 + 𝑥2/16 −9𝑥2/16 AC2 – AB2 = (𝑥2 − 9𝑥2)/16 AC2 – AB2 = −8𝑥2/16 AC2 – AB2 = −( 𝑥2)/2 2AC2 – 2AB2 = – x2 Putting BC = x 2AC2 – 2AB2 = – BC2 2AC2 + BC2 = 2AB2 2AB2 = 2AC2 + BC2 Hence proved