Question

In triangle $ABC,\angle B={90}^{\circ }$ and $D$ is the mid-point of side Be. Prove that :
$A{C}^2=A{D}^2+3C{D}^2$

Answer 1

In $\Delta ABC,$

${AB}^2+{BC}^2={AC}^2$

as $[BC=2CD]$

${AB}^2+{4CD}^2={AC}^2$ --(1)

In $\Delta ABD,$

${AD}^2={AB}^2+{BD}^2={AB}^2+{CD}^2$ --(2)

Subtracting (1) and (2)

${AC}^2-{AD}^2={AB}^2+{4CD}^2-({AB}^2+{CD}^2){AC}^2-{AD}^2={3CD}^2{AC}^2={AD}^2+{3CD}^2$

Hence proved.