Question

In an equilateral triangle ABC, D is point in side BC such that 4CD=BC . Prove that 16 AD^2=13BC^2.

Answer 1

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ABC is an equilateral triangle in which BD = BC/4

Drawn AE perpendicular to BC.

Since AE is perpendicular to BC then BE = EC = BC/2 (In equilateral triangle perpendicular 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 Δ ABC is an equilateral triangle)

BC² - BC²/4 + [BC/2 - BC/4]²

BC² - BC²/4 + BC²/16

= (16 BC² - 4 BC² + BC²)

= 13 BC²/16

So, AD² = 13 BC²/16

= 16 AD² = 13 BC²

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

ABC is an equilateral triangle in which BD = BC/4

Drawn AE perpendicular to BC.

Since AE is perpendicular to BC then BE = EC = BC/2 (In equilateral triangle perpendicular 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 Δ ABC is an equilateral triangle)

BC² - BC²/4 + [BC/2 - BC/4]²

BC² - BC²/4 + BC²/16

= (16 BC² - 4 BC² + BC²)

= 13 BC²/16

So, AD² = 13 BC²/16

= 16 AD² = 13 BC²