Question

In the given figure $\angle B$ of $△ABC$ is an a cute angle and $AD\perp BC$. Find ${AC}^2-{AB}^2-{BC}^2+2BC.BD$

Answer 1

$AD\perp BC$

find : $A{C}^2-A{B}^2-B{C}^2+2BC-BD$..........(1)

from triangle $ABD$

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

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

and from triangle $ADC$

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

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

from eq (1) and (2)

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

$D{C}^2-B{D}^2=A{C}^2-A{B}^2$........(3)

Put value of eq (3) in eq (1)

$D{C}^2-B{D}^2-B{C}^2+2BC.BD$

$\Rightarrow (BC-BD{)}^2-(BC-DC{)}^2-B{C}^2+2BC.BD$

$(\therefore DC=BC-BD;BD=BC-DC)$

$\Rightarrow B{C}^2+B{D}^2-2BC.BD-B{C}^2-D{C}^2+2BC.DC-B{C}^2+2BC.BD$

$\Rightarrow (BC-DC{)}^2+2BC.DC-D{C}^2-B{C}^2(\therefore BD=BC-DC)$

$\Rightarrow B{C}^2+D{C}^2-2BC.DC+2BC.DC-D{C}^2-B{C}^2$

$\Rightarrow 0$ hence proved.