Question

A point $D$ is on the side $BC$ of an equilateral triangle $ABC$, such that $DC=\frac{1}{4}BC.$ Prove that ${AD}^2={13CD}^2.$

Answer 1

Proof,

In $△AEC$,

$A{E}^2=A{C}^2-E{C}^2$

$A{E}^2=A{C}^2-{\left(\frac{1}{2}AC\right)}^2$$=\frac{3}{4}A{C}^2$

$\therefore AE=\frac{\sqrt{3}}{2}AC$

In $△AED$,

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

$A{D}^2={\left(\frac{\sqrt{3}}{2}AC\right)}^2+D{C}^2$ $=\frac{3}{4}(4DC{)}^2+D{C}^2$

$\therefore A{D}^2=13D{C}^2$