Precalculus method to prove $ab^*+cd^*=0$ 
Question. Suppose $a,b,c,d\in\mathbb C$ and the following equations holds:
  $$|a|^2+|b|^2=|c|^2+|d|^2=1$$
$$ac^*+bd^*=0$$
  where $x^*$ is the complex conjugate of $x$. Prove that $ab^*+cd^*=0$.

I already have a solution involving some basic knowledge of linear algebra:
Denote $X=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. It is a unitary matrix since $$XX^H=\left(\begin{matrix}aa^*+bb^*&ac^*+bd^*\\ca^*+db^*&cc^*+dd^*\end{matrix}\right)=\left(\begin{matrix}|a|^2+|b|^2&ac^*+bd^*\\(ac^*+bd^*)^*&|c|^2+|d|^2\end{matrix}\right)=I_2$$
Therefore, $$I_2=X^HX=\left(\begin{matrix}-&-\\ab^*+cd^*&-\end{matrix}\right)$$Hence the result.
Motivation
Because this question looks elementary, out of curiosity, I'm wondering if there is a method without using matrices.
I tried letting $(a,b,c,d)=(e^{i\alpha}\sin\varphi,e^{i\beta}\cos\varphi,e^{i\gamma}\sin\theta,e^{i\delta}\cos\theta)$ then plug it into the second known relation but I don't know how to simplify it.
 A: We know that
$$\begin{cases}
aa^*+bb^*=1\\
cc^*+dd^*=1\\
ac^*+bd^*=0
\end{cases}$$
Starting from the third equality we multiply it with $b^*c$ to get
$$ab^*c^*c+bb^*cd^*=0$$
now using the first two we get
$$ab^*(1-dd^*)+cd^*(1-aa^*)=0$$
so 
$$ab^*+cd^*=ab^*dd^*+aa^*cd^*$$
and inserting the third again this time conjugated ($a^*c+b^*d=0$) in the RHS we get
$$ab^*+cd^*=ab^*dd^*-ab^*dd^*=0$$
which is exactly what we were looking for.
A: For convenience, let $A=a^*$, $B=b^*$, $C=c^*$, and $D=d^*$.  We then have
$$aA+bB=1,$$
$$cC+dD=1,$$
$$aC+bD=0,$$
and
$$Ac+Bd=0.$$
First we want to show that $aA=dD$.  This follows from
\begin{align}dD&=dD\cdot 1=dD(aA+bB)=adAD+bdBD\\&=adAD+(bD)(Bd)=adAD+(-aC)(-Ac)\\&=adAD+acAC=aA(cC+dD)=aA\cdot 1=aA.\end{align}
If $d=0$, then $aA=dD=0$ so that $a=0$.  Hence $aB+cD=0B+c0=0$.  We now assume $d\ne 0$.  Therefore
\begin{align}aB+cD&=aB+c\left(\frac{dD}{d}\right)=aB+c\left(\frac{aA}{d}\right)\\&=\frac{a}{d}\left(Bd+Ac\right)=\frac{a}{d}\left(Ac+Bd\right)=\frac{a}{d}\cdot 0=0.\end{align}
From this result, it follows that 
$$(a,b,c,d)=\big(e^{i\alpha}\cos \vartheta,e^{i\beta}\sin\vartheta,e^{i\gamma}\sin\vartheta,-e^{i(-\alpha+\beta+\gamma)}\cos\vartheta),$$
where $\alpha,\beta,\gamma\in[0,2\pi)$ and $\vartheta\in[0,\pi/2]$.
