Puzzle: Finding the ratio of cevians inside triangle.

Question

Given a triangle ABC, with cevians AD, BE and CF whom intersect in P. Given are the ratios |BP|/|PE| (in the case of my puzzle this is 3/2) and |FP|/|PC| (here it is 3/4). Find the ratio |AP|/|PD|.

Now I've tried looking at ceva's and obel's theorems, but I can't find the link between the ratios of |AD|/|AB| and the ratios of the cevians. (idk if this is the path to victory)

Answer 1

suppose $\frac{BP}{PE}=b$, and $\frac{FP}{PC}=c$.

$\frac{PE}{BE}=\frac{S(ABC)}{S(APC)}$, where $S(XYZ)$ is area of triangle $XYZ$, so $$\frac{PE}{BE}+\frac{PF}{CF}+\frac{PD}{AD}=\frac{S(ABP)+S(ACP)+S(BCP)}{S(ABC)}=1.$$ $$\frac{PE}{BE}=\frac{PE}{PE+PB}=\frac1{1+\frac{PB}{PE}}=\frac{1}{1+b}$$ and $$\frac{PF}{CF}=\frac{PF}{PF+PC}=\frac{\frac{PF}{PC}}{\frac{PF}{PC}+1}=\frac{c}{c+1}$$. which means $\frac{PD}{AD}=1-\frac{1}{1+b}-\frac{c}{1+c}$, and $$\frac{AP}{PD}=\frac{AD}{PD}-1=\frac{1}{\frac{b}{1+b}-\frac{c}{1+c}}-1=\frac{(b+1)(c+1)}{b-c}-1=\frac{bc+2c+1}{b-c}$$

Answer 2

Two terms from the following sum (in the L.H.S.) $$ \begin{aligned} \frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF} &= \frac {\operatorname{Area}(PBC)} {\operatorname{Area}(ABC)} + \frac {\operatorname{Area}(PCA)} {\operatorname{Area}(BCA)} + \frac {\operatorname{Area}(PAB)} {\operatorname{Area}(CAB)} \\ &= \frac {\operatorname{Area}(ABC)} {\operatorname{Area}(ABC)} =1 \end{aligned} $$ are implicitly given, and from the third one we immediately have the answer.

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Later edit: Some more details.

• Note that for two triangles like $\Delta PBC$ and $\Delta ABC$ that share a common base, in this example $BC$, the proportion of their areas is the same as the proportion of the corresponding heights w.r.t. the common base. Drawing the heights, we see that the proportion of the heights is $PD:AD$. ($AD$ was corrected as denominator above, the denominators are full cevians, the numerators are the part of the cevians from $P$ to their side intersections $D,E,F$.)

• To see that two of the three fractions are implicitly given, let us start with the example $FP:PC=3:4$. Then $$ \frac{FP}{FC} = \frac{FP}{FP+PC} = \frac{3}{3+4}\ . $$