Question

In the given figure $DB\perp BC,DE\perp AB$ and $AC\perp BC$. Prove that $\frac{BE}{DE}=\frac{AC}{BC}$

Answer 1

$let\angle ABC=\theta -\left(1\right)$

$⟹\angle DBE=90-\theta$

$In△DBE$

$\angle BDE+\angle DEB+\angle DBE={180}^{\circ }$

$⟹\angle BDE=180-90-(90-\theta )$

$=\theta -\left(2\right)$

$In△DBE\&△ABC:$

$⟹\angle DEB=\angle ACB={90}^{\circ }\left(given\right)$

$\angle BDE=\angle ABC\left(from\right(1\left)\&\right(2\left)\right)$

$⟹△DBE\sim △ABC$

$⟹\frac{DE}{BE}=\frac{AC}{BC}$