In an equilateral - ABC - E is any point on BC such that BE -1-4 BC- Prove that 16 AE 2-13 AB 2-

Given BE = 1/4 BC Draw AD ⊥ BC Let AB = BC =AC = a BE = 1/4a [Given] AD = √(3)/2a— (i) (∵ AD is altitude) BD = 1/2a (∵ D is midpoint) ED = BD BE = 1/2a ...

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Byju's Answer

Standard X

Mathematics

Pythagoras Theorem

In an equilat...

Question

In an equilateral $Δ A B C, E$ is any point on BC such that $B E = \frac{1}{4} B C .$ Prove that $16 A E^{2} = 13 A B^{2} .$

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Solution

Given $B E = \frac{1}{4} B C$
Draw $A D ⊥ B C$
Let AB = BC =AC = a
BE = $\frac{1}{4} a$ [Given]
AD = $\frac{\sqrt{3}}{2} a$ ----(i) $(∵$ AD is altitude)

BD = $\frac{1}{2} a$ ( $∵$ D is midpoint)
ED = BD - BE
= $\frac{1}{2} a - \frac{1}{4} a$
= $\frac{1}{4} a$ ----(ii)
In right $Δ A E D$
$A E^{2} = A D^{2} + D E^{2}$
From (i) and (ii)
$A E^{2} = \frac{3 a^{2}}{4} + \frac{a^{2}}{16}$
$A E^{2} = \frac{12 a^{2} + a^{2}}{16}$
$16 A E^{2} = 13 a^{2}$
$∴ 16 A E^{2} = 13 A B^{2}$

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In an equilateral ΔABC,E is any point on BC such that BE=14BC. Prove that 16AE2=13AB2.

Given BE=14BC Draw AD⊥BC Let AB = BC =AC = a BE = 14a [Given] AD = √32a----(i) (∵ AD is altitude) BD = 12a (∵ D is midpoint) ED = BD - BE = 12a−14a =14a----(ii) In right ΔAED AE2=AD2+DE2 From (i) and (ii) AE2=3a24+a216 AE2=12a2+a216 16AE2=13a2 ∴ 16AE2=13AB2

In an equilateral $Δ A B C, E$ is any point on BC such that $B E = \frac{1}{4} B C .$ Prove that $16 A E^{2} = 13 A B^{2} .$