Question

Question 14
The perpendicular from A on side BC of a $\Delta ABC$ intersects BC at D such that DB = 3CD .Prove that $2A{B}^2=2A{C}^2+B{C}^2$.

Answer 1

Given that in $\Delta ABC$, we have
$AD\perp BCandBD=3CD$
In right angle triangles ADB and ADC, we have
$A{B}^2=A{D}^2+B{D}^2...\left(i\right)$
$A{C}^2=A{D}^2+D{C}^2...\left(ii\right)$ [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
$A{B}^2-A{C}^2=B{D}^2-D{C}^2$
$=9C{D}^2-C{D}^2[\therefore BD=3CD]$
$=8C{D}^2$
$=8{\left(\frac{BC}{4}\right)}^2[Since,BC=DB+CD=3CD+CD=4CD]$
Therefore, $A{B}^2-A{C}^2=\frac{B{C}^2}{2}$
$\Rightarrow 2(A{B}^2-A{C}^2)=B{C}^2$
$\Rightarrow 2A{B}^2-2A{C}^2=B{C}^2$
$\therefore 2A{B}^2=2A{C}^2+B{C}^2$