# Finding the ratio in which two segments divide each other

Three points $A_1$, $B_1$, $C_1$ are located on the sides $BC$, $AC$, $AB$ of the triangle ABC respectively. It is known that $BA_1:A_1C = 3:1$, $AC_1:C_1B=CB_1:B_1A = 1:5$. How can I find the ratio in which segments $AA_1$ and $B_1C_1$ divide each other? I guess I need to use the Menelaus theorem here, but I can't guess what triangles and lines to choose.

• draw a diagram. generalized angle bisector theorem should be enough to solve this applied to the two triangles that share the vertex A and sides along lines AB and AC. you can find more here: en.wikipedia.org/wiki/Angle_bisector_theorem – user625 Jun 26 '18 at 2:06

Generalized angle bisector theorem 1. Apply to $\triangle AC_1B_1$ to get: $\frac{AC_1}{AB_1} \cdot \frac{B_1X}{C_1X}$ = D, where D is the ratio of sines.

2. Apply to $\triangle ABC$ to get: $\frac{AB}{AC} \cdot \frac{CA_1}{BA_1}$ = D

3. $\frac{AC_1}{AB_1} = \frac{AB}{AC} \cdot \frac{1}{5}$ from the ratios you are given.

Thus combining 1, 2 & 3: $\frac{1}{5} \cdot \frac{B_1X}{C_1X} = \frac{CA_1}{BA_1} = \frac{1}{3}$ i.e., $\frac{B_1X}{C_1X} = \frac{5}{3}$.

Let $\vec{AB}=\vec{a}$, $\vec{AC}=\vec{b},$ $\vec{AX}=x\vec{AA_1}$ and $\vec{C_1X}=y\vec{C_1B_1}.$

Thus, $$\vec{C_1A}+\vec{AX}=\vec{C_1X}$$ or $$-\frac{1}{6}\vec{a}+x\left(\vec{a}+\frac{3}{4}(-\vec{a}+\vec{b})\right)=y\left(-\frac{1}{6}\vec{a}+\frac{5}{6}\vec{b}\right),$$ which gives $$-\frac{1}{6}+\frac{1}{4}x=-\frac{1}{6}y$$ and $$\frac{3}{4}x=\frac{5}{6}y$$,which gives $$(x,y)=\left(\frac{5}{12},\frac{3}{8}\right)$$ and $$AX:XA_1=5:7$$, $$C_1X:XB_1=3:5.$$