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