Question

In triangle ABC is an equilateral triangle point D is on side BC such that BD 1/5 BC , prove that 25AD^2 =21 AB^

Answer 1

$\textbf{Given:}$

$\text{ \triangle ABC is an equilateral triangle and BD=\frac{1}{5}BC }$

$\textbf{To prove:}$

$25\;AD^2=21\;AB^2$

$\text{Since ABC is an equilateral triangle, AB=BC=AC=x(say)}$

$\text{Draw AE \perp BC}$

$\text{Then,}AE=\frac{\sqrt{3}}{2}x$

$\text{In \triangle AEB, by pythagoras theorem}$

$AD^2=AE^2+DE^2$

$AD^2=(\frac{\sqrt{3}}{2}x)^2+(BE-BD)^2$

$AD^2=\frac{3}{4}x^2+(\frac{x}{2}-\frac{x}{5})^2$

$AD^2=\frac{3x^2}{4}+(\frac{5x-2x}{10})^2$

$AD^2=\frac{3x^2}{4}+(\frac{3x}{10})^2$

$AD^2=\frac{3x^2}{4}+\frac{9x^2}{100}$

$AD^2=\frac{75x^2}{100}+\frac{9x^2}{100}$

$AD^2=\frac{84x^2}{100}$

$AD^2=\frac{21x^2}{25}$

$AD^2=\frac{21\;AB^2}{25}$

$\implies\boxed{\bf\;25\;AD^2=21\;AB^2}$

$\textbf{Hence proved}$

Find more:

In triangle ABC angle BAC is 90 degree seg BL and seg CM are medians of triangle ABC. Then prove that 4( BL²+CM²)= 5 BC²

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Answer 2

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