Ratio of bisected cevian in triangle given intersection point

Question

I have the coordinates of points $A$ $B$ and $C$ that form triangle $\triangle ABC$, and the coordinates of a point $D$ inside of $\triangle ABC$. Imagine a cevian, connecting points $A$ and $D$, and intersecting line $BC$ at point $A'$. I need to find the ratio between the lengths of line segments $AD$ and $DA'$ using only the coordinates of points $A$, $B$, $C$, and $D$.

explanatory image

Answer 1

Since no coordinates are given, let $A=(a_{1},a_{2})$, $B=(b_{1},b_{2})$, $C=(c_{1},c_{2})$ and $D=(d_{1},d_{2}).$ Then we can find the equations of the necessary lines as follows

$$AD: \hspace{15pt}y=\frac{d_{2}-a_{2}}{d_{1}-a_{1}}(x-d_{1})+d_{2}$$ $$BC: \hspace{15pt}y=\frac{c_{2}-b_{2}}{c_{1}-b_{1}}(x-c_{1})+c_{2}.$$

To find $A'$, we need to find when these two lines intersect. So, set the two equations equal to obtain $$\frac{d_{2}-a_{2}}{d_{1}-a_{1}}(x-d_{1})+d_{2}=\frac{c_{2}-b_{2}}{c_{1}-b_{1}}(x-c_{1})+c_{2}$$ and solve for $x$. To find the $y$ coordinate, use either the formula for $AD$ or $BC$ as $A'$ must be one both of the lines.

I'll leave this for you to try. Once you find the coordinates for $A'$, simply apply the distance formula in the following way (I'll let $d(A,B)$ denote the standard distance between $A$ and $B$).

$$\text{Ratio of lengths}=\frac{d(A,D)}{d(D,A')}.$$