# Why inner product < , > on $C^n$ must satisfy the parallelogram law?

Why must norm induced by an inner product < , > on $C^n$ satisfy the parallelogram law? I know that there is a proof using $||v|| = \sqrt{(< v, v>)}$. But my concern is that why it still holds if ||v|| is defined in other ways? I mean, as far as I know norm is not not necessarily defined as $||v|| = \sqrt{(< v, v>)}$. Hence, I am wondering why proving parallelogram using $||v|| = \sqrt{(< v, v>)}$ means that parallelogram laws apply to all norms?

The parallelogram law does not apply to all norms. In fact it only applies to norms which are induced by an inner product in the sense that $\| x \| = \sqrt{\langle x,x \rangle}$. The inner product here does not need to be the Euclidean inner product, however.
Many norms do not satisfy the parallelogram law. For example, any $p$-norm for $p \neq 2$ on $\mathbb{C}^n$ does not satisfy the parallelogram law.
You have a result stating that the space is Euclidian (meaning that the norm has the form $\|X\|^2 = \langle X,X \rangle$ with $\langle . \rangle$ a scalar product) iff the parallelogram identity is true.
In these conditions, if $\|X\|^2 = \langle X,X \rangle$ then chack that $$\langle X,Y \rangle = \frac 12\left[\|X+X^2\| - \|X\| ^2 - \|Y\|^2\right]$$ is a scalar product.