Question

In $\Delta$ABC, seg AD $\perp$ seg BC , DB $=3$CD. Prove that: $2A{B}^2=2A{C}^2+B{C}^2$.

Answer 1

We are given

here $\overline{AD}\perp \overline{BC}$

& $DB=3CD---\left(1\right)$

so, $BC=DB+CD$

$BC=3CD+CD$

so, $BC=4CD---\left(2\right)$

Hence it given that, $\overline{AB}$ is perpendicular to $\overline{BC}$ by the pythagoras theorem

In both triangle $△ABD$ & $△ACD$

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

$\therefore A{D}^2=A{B}^2-D{B}^2$ & $A{D}^2=A{C}^2-C{D}^2$

Compare both $A{D}^2$ with each other we get,,

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

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

$A{B}^2=A{C}^2+(3CD{)}^2-C{D}^2$ (from $\left(1\right)$)

$A{B}^2=A{C}^2+9C{D}^2-C{D}^2$

$A{B}^2=A{C}^2+8C{D}^2$

$A{B}^2=A{C}^2+8{\left(\frac{BC}{4}\right)}^2$ (from $\left(2\right)$)

$2A{B}^2=2A{C}^2+B{C}^2$ (Let multiply equation with $2$)