Suppose $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ such that... Suppose $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ such that $\langle v,w\rangle_1=0$ if and only if $\langle v,w\rangle_2=0$. Prove that there is a positive number $c$ such that $\langle\cdot,\cdot\rangle_1=c\langle\cdot,\cdot\rangle_2$ for every $v,w\in V$.
I'm at a loss on how to start this. Any guidance is appreciated.

first part:
Assume that $$
\langle x, y \rangle_1 = \langle x, y \rangle_2 = 0 \\
\langle x, x \rangle_2 = \langle y, y \rangle_2 = 1
\langle x, x \rangle_1 = c \\
\langle y, y \rangle_1 = d \\
$$
Then:
$$
\langle x + y, x - y \rangle_2 = 0
\\ \implies 0 = \langle x + y, x - y \rangle_1
= \langle x, x \rangle_1 - \langle y, y \rangle_1 
$$
second part:
consider a finite dimensioned subspace $V'$, an 
$\langle ., . \rangle_1$-orthonormal basis 
$(e_1\dots e_d)$ if $V'$.
It is also a $\langle ., . \rangle_2$-orthogonal basis;
its matrix is, according to the first part, $ a I_d$, for some $a>0$; or:
$$
 \langle x, y \rangle_1 = a \langle x, y \rangle_2 
$$
It is true for any $V'$, hence it remains true on $V$.
