Question

In $△$ ABC, $\angle$ C is an obtuse angle. $AD\perp BC$ and ${AB}^2=C^2+{3BC}^2$. Prove that the BD = 2CD.

Answer 1

$A{B}^2=A{C}^2+3B{C}^2$

$A{B}^2=A{C}^2+3(BD-CD{)}^2$ (1)

Applying Pythagoras theorem in $△ABD$ gives $A{B}^2=B{D}^2+A{D}^2$ (2)

Applying Pythagoras theorem in $△ACD$ gives $A{C}^2=C{D}^2+A{D}^2$ (3)

Substituting (2) and(3) in (1)

$B{D}^2+A{D}^2=C{D}^2+A{D}^2+3(BD-CD{)}^2$

$B{D}^2=C{D}^2+3(BD-CD{)}^2$

$B{D}^2-C{D}^2=3(BD-CD{)}^2$

$(BD-CD)(BD+CD)=3(BD-CD{)}^2$

$BD+CD=3(BD-CD)$

$BD+CD=3BD-3CD$

$4CD=2BD$

$BD=2CD$

Hence proved.