In a parallelogram, let's call it ABCD, let points E and F be the midpoints of sides AB and BC respectively. Find the ration of AP:AF, where P is the intersection of segments AF and CE

Guest Jun 2, 2021

#4**+2 **

Ok I have simplified it and made it a 2x by 2y rectangle (which is a special case parallelogram)

This is a rough outline:

Triangle DPC and triangle EPB are similar. DC=2x and EB=x so the little one is half the size of the big one.

So JP=2*IP

so 3*IP= 2y

IP = 2y/3

so PG= y- 2y/3 = y/3

Now look at similar triangles AF and i forgot to give it a name .. . ok look at the big green triangle and the similar little grey one.

PG is y/3 and A? is y so the big one is 3 times as big as the little one.

So AP:AF = 2:3

Melody Jun 2, 2021

#3**+1 **

I don't know but i have an interactive graph here:

https://www.geogebra.org/classic/fydwxkve

And it is clear that P will lay on the diagonal BD. Maybe you can use that ... I really don't know.

Melody Jun 2, 2021

#4**+2 **

Best Answer

Ok I have simplified it and made it a 2x by 2y rectangle (which is a special case parallelogram)

This is a rough outline:

Triangle DPC and triangle EPB are similar. DC=2x and EB=x so the little one is half the size of the big one.

So JP=2*IP

so 3*IP= 2y

IP = 2y/3

so PG= y- 2y/3 = y/3

Now look at similar triangles AF and i forgot to give it a name .. . ok look at the big green triangle and the similar little grey one.

PG is y/3 and A? is y so the big one is 3 times as big as the little one.

So AP:AF = 2:3

Melody Jun 2, 2021

#6**+2 **

A vector geometry method gets the answer easily.

For a diagram, imagine Melody's rectangle 'bashed' at A so that AB remains parallel to CD, so that we do have a parallelogram.

We have

\(\displaystyle \\ \underline{AF} = \underline{AB} + \underline{BF}= \underline{AB} + (1/2)\underline{BC} \dots(1)\\ \text{and}\\ \underline{EC}=\underline{EB}+\underline{BC} = (1/2)\underline{AB}+\underline{BC} \dots(2).\)

Let

\(\displaystyle \underline{AP}=\lambda\underline{AF}, \text{ and }\underline{EP}=\mu\underline{EC}, \\ 0< \lambda <1 \text{ and } 0< \mu <1. \)

From the triangle AEP,

\(\displaystyle \underline {AP} = \underline{AE}+\underline{EP},\)

and so substituting (1) and (2),

\(\displaystyle \lambda \{\underline{AB}+(1/2)\underline{BC}\}= (1/2)\underline{AB}+\mu\{(1/2)\underline{AB}+\underline{BC}\},\)

so

\(\displaystyle \underline{AB}\{\lambda - (1/2)-(\mu/2)\}=\underline{BC}\{\mu-(\lambda/2)\}.\)

AB and BC are not parallel, so for this to be true it's necessary that

\(\displaystyle \lambda -(1/2)-(\mu/2)=0\)

and

\(\displaystyle \mu - (\lambda/2)=0, \)

from which

\(\displaystyle \lambda = 2/3 \text{ and } \mu = 1/3.\)

Guest Jun 2, 2021