Question

In an equilateral $\Delta ABC$, D is a point on side BC such that $BD=\frac{1}{3}BC$. Prove that $9{\left(AD\right)}^2=7{\left(AB\right)}^2$

Answer 1

In $△ABC$ and $△ACE$

$AB=AC$ ( Given)

$AE=AE$ ( common)

$\angle AEB=\angle AEC={90}^{\circ }$

$\therefore △ABC\cong △ACE$ ( For RHS criterion)

$BE=EC$ (By C.P.C.T)

$BE=EC=\frac{BC}{2}$

In a right angled triangle, $ADE$

$A{D}^2=A{E}^2+D{E}^2$ ---------(1)

In a right angled triangle $ABE$

$A{B}^2=A{E}^2+B{E}^2$ ---------(2)

From eqn. (1) and (2) we obtain

$⟹A{D}^2-A{B}^2=D{E}^2–B{E}^2$

$⟹A{D}^2-A{B}^2=(BE–BD{)}^2–B{E}^2$

$⟹A{D}^2–A{B}^2={(\frac{BC}{2}–\frac{BC}{3})}^2–{\left(\frac{BC}{2}\right)}^2$

$⟹A{D}^2–A{B}^2={\left(\frac{3BC–2BC}{6}\right)}^2–{\left(\frac{BC}{2}\right)}^2$

$⟹A{D}^2-A{B}^2=\frac{B{C}^2}{36}–\frac{B{C}^2}{4}$ ( In a equilateral triangle $ABC$, $AB=BC=CA$)

$⟹A{D}^2=A{B}^2+\frac{A{B}^2}{36}–\frac{A{B}^2}{4}$

$⟹A{D}^2=\frac{(36A{B}^2+A{B}^2–9A{B}^2)}{36}$

$⟹A{D}^2=\frac{28A{B}^2}{36}$

$⟹A{D}^2=\frac{7A{B}^2}{9}$

$9A{D}^2=7A{B}^2$