In an equilateral triangle ABC,D is a point on side BC such that 4BD = BC. prove that 16AD^{2} = 13BC^{2}

ABC is an equilateral triangle in which BD = BC / 4. Draw AE perpendicular to BC.

Since AE is perpendicular to BC, then BE = EC = BC / 2 [ In equilateral Δ, ⊥ drawn from vertex to base bisects the base]

In Δ AED, AD^{2}^{ }= AE^{2}^{ }+ DE^{2}^{ }[**Pythagoras Theorem**] ..............(1)

In Δ AEB, AB^{2}^{ }= AE^{2}^{ }+ BE^{2}^{ }[**Pythagoras Theorem**] ............(2)

Putting the value of AE^{2}^{ }from (2) in (1), we get

AD^{2}^{ }= AB^{2}^{ }- BE^{2} ^{ }+ DE^{2}^{ }

^{ }AD^{2}^{ }= BC^{2}^{ }- (BC / 2)^{2 }+ (BE - BD)^{2 }[ As, ABC is an equilateral Δ, AB = BC = CA ]

⇒ 16 AD^{2}^{ }= 13 BC^{2}

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