In the given figure, is a right angled triangle at ∠ = 90°. is the midpoint of . Show that AC2 = AD2 + 3CD2.

It is given that is the midpoint. Therefore, = = BC/2⇒ = 2From ∆, on using Pythagoras theorem we get2 = 2 + 22 = 2 + 4D22 = (2 + D2)+ 3D22 = AD2 + 3D2 [∵2 + 2 = 2 = ]

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In the given figure, 𝐴𝐵𝐶 is a right angled triangle at ∠𝐵 = 90°. 𝐷 is the midpoint of 𝐵𝐶. Show that AC²= AD² + 3CD².

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Detailed Solution

It is given that 𝐵𝐷 is the midpoint. Therefore,

𝐵𝐷 = 𝐶𝐷 = BC/2

⇒ 𝐵𝐶 = 2𝐵𝐷

From ∆𝐴𝐵𝐶, on using Pythagoras theorem we get

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

𝐴𝐶² = 𝐴𝐵² + 4𝐵D²

𝐴𝐶² = (𝐴𝐵² + 𝐵D²)+ 3𝐵D²

𝐴𝐶² = AD²+ 3𝐵D²[∵𝐴𝐵² + 𝐵𝐷² = 𝐴𝐷² 𝑎𝑛𝑑 𝐵𝐷 = 𝐶𝐷]

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