Question

In the above figure given of $△ABC,\angle ACB={90}^{\circ }$

D is the midpoint of BC

Prove that ${AB}^2$= ${4AD}^2-{3AC}^2$

Answer 1

In $△ABC,\angle C={90}^{\circ }$

$BD=CD=\frac{BC}{2}$

By Pythagoras Theorem,

$(Hypotenuse{)}^2=(Base{)}^2+(Height{)}^2$

In $△ABC;A{B}^2=B{C}^2+A{C}^2...\left(1\right)$

In $△ADC;A{D}^2=A{C}^2+D{C}^2...\left(2\right)$

From (1) $A{B}^2=B{C}^2+A{C}^2$

$A{B}^2=(2CD{)}^2+A{C}^2$ $(\because BC=2CD)$

$A{B}^2-A{C}^2=4C{D}^2$

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

Substituting $C{D}^2$ value from (3) in (2)

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

$4A{D}^2=4A{C}^2+A{B}^2-A{C}^2$

$4A{D}^2-4A{C}^2+A{C}^2=A{B}^2$

$A{B}^2=4A{D}^2-3A{C}^2$

Hence proved