Question

State true or false:

In an acute-angled $△ABC$, $AD$ is perpendicular to side $BC$, then $A{C}^2=A{B}^2+B{C}^2-2BC\times BD$

• A. True
• B. False

Answer 1

The correct option is A True

Given: Acute angled $△ABC$, $AD\perp BC$,
Now, in $△ADB$,
$A{B}^2=A{D}^2+B{D}^2$ (Pythagoras theorem)
$\Rightarrow A{D}^2=A{B}^2-B{D}^2$ .....(I)
In $△ADC$,
$A{C}^2=A{D}^2+D{C}^2$
$\Rightarrow A{D}^2=A{C}^2-C{D}^2$ ....(II)
Equating (I) and (II), we get
$A{B}^2-B{D}^2=A{C}^2-C{D}^2$
$A{C}^2=A{B}^2+C{D}^2-B{D}^2$
$A{C}^2=A{B}^2+(BC-BD{)}^2-B{D}^2$ .....($BD+DC=BC$)
$A{C}^2=A{B}^2+B{C}^2+B{D}^2-2BC\times BD-B{D}^2$
$A{C}^2=A{B}^2+B{C}^2-2BC\times BD$