Question 67
Let $ABCD$ be a cyclic quadrilateral with $\hat{B} \neq 90^\circ$ and
$$
AB^2 + BC^2 + CD^2 + DA^2 = 2AC^2.
$$
Prove that the midpoint of diagonal $BD$ is on $AC$.
$$
AB^2 + BC^2 + CD^2 + DA^2 = 2AC^2.
$$
Prove that the midpoint of diagonal $BD$ is on $AC$.