Problem - 3315
In $\triangle{ABC}$, let $a$, $b$, and $c$ be the lengths of sides opposite to $\angle{A}$, $\angle{B}$ and $\angle{C}$, respectively. $D$ is a point on side $AB$ satisfying $BC=DC$. If $AD=d$, show that
$$c+d=2\cdot b\cdot\cos{A}\quad\text{and}\quad c\cdot d = b^2-a^2$$
Applying Law of Cosines on $\triangle{ABC}$ and $\triangle{ADC}$:
$$a^2 = b^2 +c^2 - 2\cdot b\cdot c\cdot\cos{A}\quad\text{and}\quad a^2 = b^2 +d^2 - 2\cdot b\cdot d\cdot\cos{A}$$
Rearranging these two equations leads to:
$$c^2 - 2\cdot b\cdot c\cdot\cos{A} + b^2 - a^2 = 0$$
$$d^2 - 2\cdot b\cdot d\cdot\cos{A} +b^2-a^2=0$$
Clearly $c > d$, therefore they are the two roots of the quadratic equation:$$x^2-2\cdot b\cdot\cos{A} +(b^2-a^2)=0$$
Hence, by Vieta's theorem, the following must hold:
$$c+d = 2\cdot b\cdot \cos{A}\quad\text{and}\quad b\cdot d=b^2-a^2$$