Showing the sum of orthogonal projections with orthogonal ranges is also an orthogonal projection 
Show that if $P$ and $Q$ are two orthogonal projections with orthogonal ranges, then $P+Q$ is also an orthogonal projection.

First I need to show $(P+Q)^\ast = P+Q$. I am thinking that since
\begin{align*}
((P+Q)^\ast f , g) & = (f,(P+Q)g) \\
 & = (f,Pg) + (f,Qg) \\
 & = (P^\ast f,g) + (Q^\ast f,g) \\
 & = (Pf,g) + (Qf,g) \\
 & = ((P+Q)f,g),
\end{align*}
we get $(P+Q)^\ast=P+Q$.
I am not sure if what I am thinking is right since I assumed that $(P+Q)f=Pf+Qf$ is true  for any bounded linear operator $P$, $Q$.
For $(P+Q)^2=P+Q$, I use
$$(P+Q)^2= P^2 + Q^2 + PQ +QP,$$
but I cant show $PQ=0$ and $QP=0$. 
Anyone can help me? Thanks.

To complete your proof we need the following observations. 
If $\langle f,g\rangle=0$ for all $g\in H$, then $f=0$. Indeed, take $g=f$, then you get $\langle f,g\rangle=0$. By definition of inner product this implies $f=0$.
Since $\mathrm{Im}(P)\perp\mathrm{Im}(Q)$, then for all $f,g\in H$ we have $\langle Pf,Qg\rangle=0$. which is equivalent to $\langle Q^*Pf,g\rangle=0$ for all $f,g\in H$. Using observation from previous section we see that $Q^*P(f)=0$ for all $f\in H$, i.e. $Q^*P=0$. Since $Q^*=Q$ and $P^*=P$ we conclude 
$$
QP=Q^*P=0
$$
$$
PQ=P^*Q^*=(QP)^*=0^*=0
$$
In fact $R$ is an orthogonal projection iff $R=R^*=R^2$. In this case we can prove your result almost algebraically
$$
(P+Q)^*=P^*+Q^*=P+Q
$$
$$
(P+Q)^2=P^2+PQ+QP+Q^2=P+0+0+Q=P+Q
$$
I said almost, because prove of $PQ=QP=0$ requires some machinery with elements of $H$ and its inner product.
