I think first one is $1$ and the second one is $1/3$. Again by using Menelaus theorem

By Mass points, $$A=1 , B=2 , C=1 , D=3 , E=2 , P=4$$Hence, $BP:PE=1$ and $AP:PD=3$

AlastorMoody wrote: By Mass points, $$A=1 , B=2 , C=1 , D=3 , E=2 , P=4$$Hence, $BP:PE=1$ and $AP:PD=3$ Is this allowed in rmo...the mass point techniques..(u will.learn later the com)

Pigeonhole_Biswaranjan_ wrote: AlastorMoody wrote: By Mass points, $$A=1 , B=2 , C=1 , D=3 , E=2 , P=4$$Hence, $BP:PE=1$ and $AP:PD=3$ Is this allowed in rmo...the mass point techniques..(u will.learn later the com) I didn't get it?? Wht do you mean ??

Anybody with a nice solution? I am sorry for the long bump but anybody?

Let $F$ be the midpoint of $DC$, so that $D, F$ are points of trisection of $BC$. Now in triangle $CAD$, $F$ is the mid-point of $CD$ and $E$ is that of $CA$. Hence $\frac{CF}{FD} =1=\frac{CE}{EA}$. Thus $EF \parallel AD$. Hence we find that $EF \parallel PD$. Hence $\frac{BP}{PE} = \frac{BD}{DF}$. But $BD = DF$. We obtain $\frac{BP}{PE}= 1$. In triangle $ACD$, since $EF \parallel AD$ we get $\frac{EF}{AD} = \frac{CF}{CD} = \frac{1}{2}$. Thus $AD = 2EF$. But $\frac{PD}{EF} = \frac{BD}{BF} = \frac{1}{2}$. Hence $EF = 2PD$. Therefore $AD = 2EF = 4PD$. This gives $AP = AD$ - $PD= 3PD$. We obtain $\frac{AP}{PD} = 3$. This is the official solution.

Just construct $EX \parallel AD$ and you are done.

The way I've done it - Construct G midpoint of BD. (i) So, by midpoint theorem, EG||AD, so EG||PD. And since 2BD = DC, BD = DF. So, in triangle AEG, by midpoint theorem, BP = PD. So, BP/PD = 1. (ii) Now, Let EG = x. So, PD = x/2, AD = 2x. So, AP = 3x/2. Thus, AP/PD = 3. P.S. If someone could tell me how to write latex that would be great!

Pi_Person00 wrote: The way I've done it - Construct $G$ midpoint of $BD$. (i) So, by midpoint theorem, $EG \parallel AD,$ so $EG \parallel PD$. And since $2BD = DC,$ $BD = DF$. So, in $\triangle AEG,$ by midpoint theorem, $BP = PD$. So, $\frac{BP}{PD} = 1$. (ii) Now, let $EG = x$. So, $PD = \frac{x}{2}, AD = 2x$. So, $AP = \frac{3x}{2}$. Thus, $\frac{AP}{PD} = 3$. FTFY, and to write $\LaTeX$ just memorize a bunch of commands

rg_ryse wrote: Pi_Person00 wrote: The way I've done it - Construct $G$ midpoint of $BD$. (i) So, by midpoint theorem, $EG \parallel AD,$ so $EG \parallel PD$. And since $2BD = DC,$ $BD = DF$. So, in $\triangle AEG,$ by midpoint theorem, $BP = PD$. So, $\frac{BP}{PD} = 1$. (ii) Now, let $EG = x$. So, $PD = \frac{x}{2}, AD = 2x$. So, $AP = \frac{3x}{2}$. Thus, $\frac{AP}{PD} = 3$. FTFY, and to write $\LaTeX$ just memorize a bunch of commands Thanks, but I need a proper website.

Learn latex here: https://artofproblemsolving.com/wiki/index.php/LaTeX

Here is another solution. Use barycentrics on $\triangle ABC$. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1),D=(2:1:0),E=(1:0:1)$. Since $BE$ and $AD$ are cevians, $P=(2:1:2)$. Normalizing everything, we seek $\frac{BP}{PE}=\frac{[ABP]}{[APE]}$ and $\frac{AP}{PD}=\frac{[ABP]}{[PBD]}$ which are immediate by the area formula.

Hello this is mine solution : we proceed by BDT bashing call [x] as the area of the closed figure x then let $[APE]=\delta$ as in triangle APC : PE is the median then $[APE]=[EPC]=\delta$ $\cdots$ call $[BPD]=\beta$ then by BDT in triangle $BPC : [PDC]=2\beta$ now call [ABP]=X then by BDT in triangle $$ABC : \frac{X+\beta}{2\delta + 2\beta} = \frac{1}{2}$$which gives $X=\delta$ now again by BE as median in trinagle ABC we get $$2\delta = 3\beta + \delta$$which means $\frac{\delta}{\beta}=3$ which means $\frac{AP}{PD} = 3$ and $\frac{BP}{PE}=1$ by BDT in triangle ABD and ABE respectively

use mid point theorm and similarity

Attachments:

Construct $EX \parallel AD$, and $DY \parallel BE$ Note that, by midpoint theorem in $\triangle CDA$, $X$ is the midpoint of $CD$, which gives $BD=DX=CD$. Now by basic proportionality theorem in $\triangle BDP$ and $\triangle BXE$, we have that $\frac{BP}{PE}=\frac{BD}{DX}=1$. Now we use BPT in $\triangle CDY$ and $\triangle CBE$ $\frac{CD}{DB}=\frac{CY}{YE}=\frac{2}{3}$. Thus, $EY=\frac{CE}{3}=\frac{AE}{3}$. Again, using BPT in $\triangle APE$ and $\triangle ADY$, we get that $\frac{AP}{AD}=\frac{AE}{EY}=3$. We are thus done.