A triangle $ABC$ is given with $AB=8,BC=9,AC=13$. The vertex $A$ lies on the plane $\alpha$ and $BC\parallel \alpha$. A line $p$ through the midpoint of $AC$ and $B$ intersects $\alpha$ in $D$. Find the length of $BD$.
$BD=11$
I am new to solid geometry and it's difficult for me to imagine the figures. I am really used to plane geometry. Can you help me with the diagram? I am not sure if we can use these formulas here, but if $K$ is the midpoint of $AC$, then $BK$ is a median of the triangle $ABC$, so $$m_b=\dfrac{1}{2}\sqrt{2a^2+2c^2-a^2}=\dfrac{11}{2}.$$ What can I do from here?
We know the plane $\alpha$ is parallel to $BC$ and so the line $AD$ is parallel to $BC$ (line $AD$ is in the same plane as $\triangle ABC$).
If $K$ is the midpoint, $\triangle AKD \cong \triangle CKB$ given $BC \parallel AD$ and $CK = AK$.
So $BD = 2 \cdot BK = 11$
EDIT: On your question on why $BC \parallel AD$, say $\triangle ABC$ is in plane $\beta$. The key point is that the line through points $A$ and $D$ are in both planes $\alpha$ and $\beta$.
Now it comes down to using the following properties of Euclidean space - (i) If a line and a plane are parallel, they never meet. As line through $B, C$ and plane $\alpha$ are parallel, they never meet and this implies lines through $B, C$ and through $A, D$ never meet (ii) If two lines in the same plane never meet, they are parallel. So lines through $B, C$ and through $A, D$ being in the same plane $\beta$ must be parallel.