Question

In an equilateral $△ABC$, the side $BC$ is trisected at $P$. Prove that $9A{P}^2=7A{B}^2$

Answer 1

$P$ is one of trisection points of $BC.$

Let $Q$ be the other.

Join $AQ$

$⟹BP=PQ=QC=\frac{BC}{3}=\frac{AB}{3}$

$In△ABQ$

$AP$ is median$[BP=PQ]$

$\therefore$ By Apollonius thm

${AB}^2+{AQ}^2=2({AP}^2+{BP}^2)$

${AB}^2+{AQ}^2=2{AP}^2+2{\left(\frac{AB}{3}\right)}^2$

$⟹\frac{7{AB}^2}{9}=2{AP}^2-{AQ}^2-\left(1\right)$

$In△AQC$

$AQ$ is median $[PQ=QC]$

By Apollonius Thm

${AC}^2+{AP}^2=2({AQ}^2+{PQ}^2)$

$⟹{AB}^2+{AP}^2=2{AQ}^2+2{\left(\frac{AB}{3}\right)}^2[AB=AC]$

$⟹\frac{7{AB}^2}{9}=2{AQ}^2-{AP}^2-\left(2\right)$

$\left(1\right)=\left(2\right)$

$⟹2{AP}^2-{AQ}^2=2{AQ}^2-{AP}^2$

$⟹3{AP}^2=3{AQ}^2$

$⟹AP=AQ$

Putting$AP=AQin\left(2\right)$

$\frac{7}{9}{\left(AB\right)}^2=2{AP}^2-{AP}^2$

$⟹7{AB}^2=9{AP}^2$