The triangle ABC has the length of the side AB equal to an integer greater than or equal to 1, and AD is the bisector of the angle BAC, D belonging to BC. It is known that the area of ABC = 3* the area of ABD. How many natural values can the length of side BC take?
MY IDEAS
MY DRAWING
Okey, so I thought of writing the both area as:
$A(ABC)=\frac {h\cdot BC}{2}=3\cdot A(ABD)= \frac{h\cdot BD}{2}$
where h= the height of BC.
$\frac {h\cdot BC}{2}=\frac{h\cdot BD}{2}$ we can multiply it by 2 and eliminate the height
$BC=3BD$
so, $DC=2BD$, then $\frac{BD}{DC}=\frac{1}{2}$
we can apply the theorm of the bisector and
$\frac{BD}{DC}=\frac{1}{2}=\frac{AB}{AC}$
After this we can write AB as a and AC as 2a
I don't know what to do along! Hope one of you can help me! Thank you!