1. ## Proved that

2. Hello, dapore!

In any triangle $\displaystyle ABC$ with angles $\displaystyle A,B,C$ and sides $\displaystyle a,b,c\!:$

. . . $\displaystyle b\cos A + a\cos B \:=\:c$
Code:
              C
o
*| *
* |   *
*  |     *
b *   |       * a
*    |         *
*     |           *
*      |             *
A o   *   o   *   *   *   o B
x   D       y
: - - - - - c - - - - - :

Construct altitude $\displaystyle CD$ to side $\displaystyle AB.$
Let $\displaystyle x = AD,\;Y = DB$

In right triangle $\displaystyle CDA\!:\;\;\cos A \,=\,\frac{x}{b} \quad\Rightarrow\quad x \,=\,b\cos A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\cos B \,=\,\frac{y}{a} \quad\Rightarrow\quad y \,=\,a\cos B$ .[2]

Note that: .$\displaystyle x + y \:=\:c$ .[3]

Substitute [1] and [2] into [3]: .$\displaystyle b\cos A + a\cos B \;=\;c$

3. Originally Posted by Soroban
Hello, dapore!

Code:
              C
o
*| *
* |   *
*  |     *
b *   |       * a
*    |         *
*     |           *
*      |             *
A o   *   o   *   *   *   o B
x   D       y
: - - - - - c - - - - - :
Construct altitude $\displaystyle CD$ to side $\displaystyle AB.$
Let $\displaystyle x = AD,\;Y = DB$

In right triangle $\displaystyle CDA\!:\;\;\cos A \,=\,\frac{x}{b} \quad\Rightarrow\quad x \,=\,b\cos A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\cos B \,=\,\frac{y}{a} \quad\Rightarrow\quad y \,=\,a\cos B$ .[2]

Note that: .$\displaystyle x + y \:=\:c$ .[3]

Substitute [1] and [2] into [3]: .$\displaystyle b\cos A + a\cos B \;=\;c$
Thanks Soroban
just for you