If $a \equiv b \pmod n$ and $c+d = n$, does $ca+bd \equiv 0 \pmod n$?

Question

I am trying to prove a different equation and am able to if the following is true, but I am not exactly sure if it is true.

If $a\equiv b \pmod{9}$ and $c+d = 9$, is $ca+bd \equiv 0 \pmod{9}$ a true statement?

I have tried various examples, like

$a = 29$, $b = 2$, $c = 2$, and $d = 7$

$a = 13$, $b = 4$, $c = 4$, and $d = 5$

and more.

If anyone has a counterexample, please let me know! Otherwise, if this is true, then can someone please explain why? Thanks!

Answer 1

If $a\equiv b\pmod n$, then this is just $$a(c+d) \equiv an\equiv 0\pmod n$$

Answer 2

$a\equiv b\pmod{9}$ and $c+d=9$ implies $a-b\equiv 0\pmod{9}$ and $d\equiv -c \pmod{9}$ ,respectively. So

$$ca+bd\equiv ca-cb\equiv c(a-b)\equiv c\cdot 0\equiv 0 \pmod{9}.$$

Answer 3

$$ \begin{align}&\ \ \ \ \ \color{#c00}a\ \equiv\,\ \color{#c00}b\\ &\ \ \ \ \ \color{#0a0}c\ \equiv \color{#0a0}{-d}_{\phantom{|}}\\ \hline {\Longrightarrow}\ \ &\ \ \ \color{#0a0}c\ \color{#c00}a\,+\,d\,b\\[.2em] {\equiv}\ \ \ & \color{#0a0}{-d}\,\color{#c00}b\,+\,d\,b\,\equiv\, 0\end{align}\qquad$$

Answer 4

Just to list another approach: $$ ac+bd = a\underbrace{(c+d)}_{\text{multiple of $9$}} + \underbrace{(b-a)}_{\text{multiple of $9$}}d, $$ (this is an equal sign, not a congruent sign) so the left-hand side must be a multiple of $9$, too.