Question 57
Let $ABC$ be a triangle in which $AB = AC$. Let $D$ be the mid-point of $BC$ and $P$ be a point on $AD$. Suppose $E$ is the foot of perpendicular from $P$ on $AC$. If $\frac{AP}{PD} = \frac{BP}{PE} = \lambda$, $\frac{BD}{AD} = m$ and $z = m^2(1+\lambda)$, prove that
$$
z^2 - (\lambda^3 - \lambda^2 - 2)z + 1 = 0
$$
Hence show that $\lambda \geq 2$ and $\lambda = 2$ if and only if $ABC$ is equilateral.
$$
z^2 - (\lambda^3 - \lambda^2 - 2)z + 1 = 0
$$
Hence show that $\lambda \geq 2$ and $\lambda = 2$ if and only if $ABC$ is equilateral.