Question

ABC is an equilateral triangle. If D is a point on side BC such that BD : BC = 1 : 3, then find $A{D}^2:A{B}^2$.

Answer 1

Given: ABC is an equilateral triangle.
BD = $\frac{1}{3}$BC = $\frac{1}{3}$AB
In $△ABC$,
$BE=\frac{1}{2}BC=\frac{1}{2}AB$
$A{E}^2=A{B}^2-B{E}^2$
$\Rightarrow$ $A{E}^2=A{B}^2-(\frac{1}{2}AB{)}^2$
$\Rightarrow$ $A{E}^2=A{B}^2-\frac{1}{4}A{B}^2$
$\Rightarrow$ $A{E}^2=\frac{3}{4}A{B}^2$ ---(1)

In $△ADE$,
$A{E}^2=A{D}^2-D{E}^2$
$\Rightarrow$ $A{E}^2=A{D}^2-(BE-BD{)}^2$
$\Rightarrow$ $A{E}^2=A{D}^2-(\frac{1}{2}AB-\frac{1}{3}AB{)}^2$
$\Rightarrow$ $A{E}^2=A{D}^2-(\frac{1}{6}AB{)}^2$
$\Rightarrow$ $A{E}^2=A{D}^2-\frac{1}{36}A{B}^2$ ---(2)

From (1) and (2),
$\frac{3}{4}A{B}^2$ = $A{D}^2-\frac{1}{36}A{B}^2$
$\Rightarrow$ $A{D}^2=\frac{3}{4}A{B}^2+\frac{1}{36}A{B}^2$
$\Rightarrow$ $A{D}^2=A{B}^2\times \left(\frac{27+1}{36}\right)$
$\Rightarrow$ $A{D}^2=\frac{28}{36}A{B}^2$
$\Rightarrow$ $A{D}^2=\frac{7}{9}A{B}^2$
$A{D}^2:A{B}^2$ = 7 : 9