In triangle ABC, D is the mid-point of BC and E is the mid-point of AD

Question

Figure not drawn to scale. https://gmatclub.com/forum/download/file.php?id=35363 In triangle ABC, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF : FC ? (A) 1:1 (B) 1:2 (C) 1:3 (D) 2:3 (E) 3:4 main-qimg-5456f088c373424b9e0b766a7449d264.png

KarishmaB · Accepted Answer

This question can be solved using the concept of ratio of intercepts on transversals is the same. So if we have have two transversals on 3 parallel lines, the ratio of their intercepts will be the same. Imagine a line parallel to BF passing through D. Say, there are another two lines parallel to these two lines passing through A and C respectively. So we have 4 parallel lines and two transversals AD and AC. Temp2.jpeg Considering 3 parallel lines: the dashed line, BF and the red line, AE/ED = AF/FP = 1/1 Considering 3 parallel lines: BF, the red line and the yellow line, BD/DC = FP/PC = 1/1 From these, we get AF = FP = PC and hence AF:FC = 1:2

BrentGMATPrepNow · Answer

VERY tricky question. I'd say it's skirting the 800 level of difficulty!

Here's a solution with some diagrams.

First, we'll add the given information.
 https://s11.postimg.cc/n16g01jbj/image.jpg

Next, from point D let's add a line that's PARALLEL to BF
 https://s11.postimg.cc/8n7nm1mvj/image.jpg

This addition of a parallel line allows us to see a variety of equivalent angles (and similar triangles!)
 https://s11.postimg.cc/4pkdwn027/image.jpg

First, let's examine these two similar triangles. 
 https://s11.postimg.cc/e7k4wolqn/image.jpg 
Since side AD (on the blue triangle) is twice the length of side AE (on the red triangle), we can conclude that the blue triangle is  twice as large  as the red triangle. 
So, side AP is twice the length of side AF. 
This tells us that   side PF = side AF

Now let's examine these two similar triangles.
 https://s11.postimg.cc/z5qaurllb/image.jpg 
Since side BC (on the purple triangle) is twice the length of side DC (on the green triangle), we can conclude that the purple triangle is  twice as large  as the green triangle. 
So, side FC is twice the length of side PC. 
This tells us that   side PF = side PC

At this point, we know that   side PF = side AF   and   side PF = side PC  
Since both equations are set equal to PF, we can conclude that   AF = FP = PC

So, the ratio of AF : FC = 1:2

Answer: B

Cheers,
Brent

laxpro2001 · Answer

This is a tough one. I guessed - the only progress I was able to make is that BD and DC are equal. I thought AD bisects angle A but I don't see how I can use that to help me and I'm not sure I can make that conclusion even though it makes sense given that BD = DC because each point on the bisector is equidistant from each the sides of the angle.

We also know that AE = ED because E is the midpoint of AD but I don't see how we can use that with what we know about BF to progress. Would love it if someone could provide a good explanation.

MaverickTone · Answer

G is on AC so as EG // BC,

-> GC = 3 FG, also DG = GH.

-> AF = 2FG

-> AF = 1/2 FC

That was my train of thought. Normally when I see mid point I always think of drawing parallels or flipping objects.

Since EG // DC => G is the midpoint of AC. or AG = GC and, EG = 1/2 DC = 1/4 BC.

Also EG // BC => EF = 1/4 BC => EF = 1/3 EB => FG = 1/3 GC

Also, G is mid point of AC, D is mid point of BC => GD // AB => GC = AG.

So AF + FG = GC and FG = 1/3 GC => Af = 2/3 GC = 2/3 AG.

FC = FG + GC = FG + AG = 1/2 AF + 3/2 AF = 2 AF.

laxpro2001 · Answer

-> GC = 3 FG, also DG = GH 
Where is H? I assume that was a typo but I cannot figure out which letter you meant.

And how do you know GC = 3FG?

Sahilkm · Answer

Hpw do I explain this without figure. Okay, let me try:

Consider a point, X, on AC somewhere between F and C. The point X should be such that DX is parallel to BF. 
Now, since BF || DX so, CX=XF (parallel base theorem, since BD=DC)
Also, AF= FX (parallel base theorem, since AE=ED)

So CX=XF=AF

Hence, AF: FC = 1:2

MaverickTone · Answer

Disregard the H, it was a typo.

Anyway, CG = 3 FG is because FG // BC, and the ratio between FG and BC is the same as FG and FC. FC = 4 FG so GC = 3FG.

Now I realize my explanation is a bit messy, but I hope it helps.

Anurag Nayak · Answer

As per the Menelaus Theorem,
(BC * ED * AF)/(BD*AE*FC)=1,

Assume BD=CD=x ; AE=ED=y ;
So, AF/FC=1/2.

laxpro2001 · Answer

Bunuel 
I can't seem to find any good resources that help teach this concept - do you know of any?

shashankism · Answer

Lets draw a line DG parallel to BF.
Now triangle GCD ~ triangle FCB
-> GC/GF = CD/BD = 1 (Since D is the midpoint of BC) 
-> GC = GF

Also triangle AFE ~ AGD
-> AF/GF = AE/ED =1 (Since E is the midpoint of AD)
-> AF = GF

--> AF = GF = GC

So AF / FC = 1/2

Answer B

KarishmaB · Answer

Check this out: https://www.youtube.com/watch?v=y5yxL24sRHI

Crytiocanalyst · Answer

Figure not drawn to scale.
 https://gmatclub.com/forum/download/file.php?id=35363 
In triangle ABC, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF : FC ?

(A) 1:1

(B) 1:2

(C) 1:3

(D) 2:3

(E) 3:4

it's pretty hard to showcase to show without diagram

when we draw line parallel to BF through D we get 2 similar triangles and when we compare and add all the sides

we get AF/FC = 1:2 
Therefore IMO B

bumpbot · Answer

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In triangle ABC, D is the mid-point of BC and E is the mid-point of AD

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In triangle ABC, D is the mid-point of BC and E is the mid-point of AD

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