Question

In fig., $ABC$ is a triangle in which $\angle ABC>{90}^o$ and $AD\perp CB$ produced. Prove that $A{C}^2=A{B}^2+B{C}^2+2BC.BD$.

Answer 1

To Prove: $A{C}^2=A{B}^2+B{C}^2+2BC.BD$

Proof:

In $△ADC$, we have

By Pythagoras Theorem,

$A{C}^2=A{D}^2+D{C}^2.....\left(1\right)$

In $△ADB$, we have

By Pythagoras Theorem,

$A{B}^2=A{D}^2+B{D}^2.......\left(2\right)$

Subtracting 2) from 1), we get

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

$A{C}^2-A{B}^2=D{C}^2-(DC-BC{)}^2$ [$\because BD=DC-BC$]

$A{C}^2-A{B}^2=D{C}^2-[D{C}^2-2DC.BC+B{C}^2]$ [Using the identity $(a-b^2=a^2-2ab+b^2)$]

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

$A{C}^2-A{B}^2=2(DB+BC)BC-B{C}^2$ [$\because DC=DB+BC$]

$A{C}^2-A{B}^2=2DB.BC+2B{C}^2-B{C}^2$

$\therefore$ $A{C}^2=A{B}^2+2DB.BC+B{C}^2$ [hence proved]