In an equilateral triangle ABC,D is a point on side BC such that 4BD = BC. prove that 16AD2 = 13BC2

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, AD2 = AE2 + DE2  [Pythagoras Theorem]  ..............(1)

In Δ AEB, AB2 = AE2 + BE2  [Pythagoras Theorem]  ............(2)

Putting the value of AE2 from (2) in (1), we get

 AD2 = AB2 - BE2 + DE2  

 AD2 = BC2 - (BC / 2)2 + (BE - BD)2  [ As, ABC is an equilateral Δ, AB = BC = CA ]

⇒ 16 AD2 = 13 BC2

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 DRAW A PERPENDICULAR ON SIDE BC NAME THE POINT E

SO, BE = EC ( ALTITUDE ACT AS A MEDIAN N AN EQUILATERAL )

AE = (a)( ROOT 3/2 ) ( AE is the altitude)

BD = BC/4 (GIVEN)

DE = BC/4

NOW by pythagoras theorem

AD2 = AE2 + DE2

substitutin value

AD2 = (aroot3/2)+ (BC/4)2

AD= 3a2/4 + BC2/16

AD2 = (12a2 + BC)/16

16AD2 = 12a+ BC2 (BC = AB = AC = a)

16AD2 = 13BC2

Hence proved

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