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# Geometry problem

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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

Jun 2, 2021

#4
+115975
+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

Jun 2, 2021

#1
+865
+1

Create similar triangles

#2
+1

I do not think we can use similar triangles here, we aren't given enough information to make an assumption.

Guest Jun 2, 2021
#3
+115975
+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.

Jun 2, 2021
#4
+115975
+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
#5
+121096
0

Very nice, Melody......I could  prove this with coordinates, but  your solution is more elegant  (I just didn't draw  enough lines.....LOL!!!!)

CPhill  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

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

Jun 2, 2021