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Summation by Sigma Method

In an equilat...

Question

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

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Solution

Let A be origin
By section formula,
$\to E = \frac{\to c + 3 \to b}{4}$
As ABC is an equilateral triangle,
$∣ ∣ \to b ∣ ∣ = | \to c |$
Now $16 A E^{2} - 13 A B^{2} = \frac{16 {∣ ∣ \to c + 3 \to b ∣ ∣}^{2}}{16} - | 3 | ∣ ∣ \to b ∣ ∣$
$= (\to c + 3 \to b) . (\to c + 3 \to b) - | 3 | {∣ ∣ \to b ∣ ∣}^{2}$
$= {| \to c |}^{2} + 9 {∣ ∣ \to b ∣ ∣}^{2} + 6 \to c . \to b - | 3 | {∣ ∣ \to b ∣ ∣}^{2}$
$= {∣ ∣ \to b ∣ ∣}^{2} + 9 {∣ ∣ \to b ∣ ∣}^{2} + \frac{6 {∣ ∣ \to b ∣ ∣}^{2}}{2} - 13 {∣ ∣ \to b ∣ ∣}^{2}$
$= 0$

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