Question

ABC is a right angled triangle. AD is perpendicular to the hypotenuse BC. If $AC=2AB$, then the value of BD is

• A. $\frac{BC}{2}$
• B. $\frac{BC}{3}$
• C. $\frac{BC}{4}$
• D. $\frac{BC}{5}$

Answer 1

The correct option is D $\frac{BC}{5}$

$B{C}^2=A{B}^2+A{C}^2$

$B{C}^2=A{B}^2+(2AB{)}^2$

$=5A{B}^2$

$BC=\sqrt{5A{B}^2}$

$A{B}^2=\frac{B{C}^2}{5}$

$A{C}^2=A{D}^2+C{D}^2$

$A{B}^2=A{D}^2+B{D}^2$

$A{C}^2-C{D}^2=A{B}^2-B{D}^2$

$B{D}^2=A{B}^2-A{C}^2+C{D}^2$

$=A{B}^2-4A{B}^2+(BC-BD{)}^2$

$B{D}^2=A{B}^2-4A{B}^2+B{C}^2+B{D}^2-2BC\cdot BD$

$2BC\cdot BD=-3A{B}^2+5A{B}^2$

$2BC\ddot{B}D=2A{B}^2$

$BD={\frac{AB}{BC}}^2=\frac{BC/5}{BC}=\frac{BC}{5}$

$\Rightarrow BD=\frac{BC}{5}$.