Question

In the given figure, $AD\perp BC$ and $BD=$$\frac{1}{3}$$CD$. Prove that $2C{A}^2=2A{B}^2+B{C}^2.$

Answer 1

Given ABC is a triangle and $AD\perp BC.$

To prove : $2C{A}^2-2.A{B}^2+B{C}^2$

proof : $BD+CD=BC$

$\frac{1}{3}.CD+CD=BC$

$CD+3CD=3.BC$

$4CD=3BC$

$CD=\frac{3}{4}.BC\cdots \left(1\right)$

From (1)

$BD=\frac{3}{4}.BC$

$BD=\frac{BC}{4}\cdots \left(2\right)$

In $△ACD,$

$A{C}^2=A{D}^2+D{C}^2\cdots \left(3\right)$

In $△ADB,$

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

$A{D}^2=A{B}^2-B{D}^2\cdots \left(4\right)$

put (4) in (1)

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

$A{C}^2=A{B}^2-{\left(\frac{BC}{4}\right)}^2+{(\frac{3}{4}.BC)}^2$

$A{C}^2=A{B}^2-\frac{B{C}^2}{16}+\frac{9.B{C}^2}{16}$

$A{C}^2=A{B}^2+\frac{-B{C}^2+9B{C}^2}{16}$

$A{C}^2=A{B}^2+\frac{8.B{C}^2}{16}$

$A{C}^2=A{C}^2=\frac{2.A{B}^2+B{C}^2}{2}$

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

Hence it proved.