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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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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
Bunuel wrote:
Figure not drawn to scale.

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

Attachment:
main-qimg-5456f088c373424b9e0b766a7449d264.png

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.
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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.
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geometry.png [ 34.55 KiB | Viewed 126726 times ]

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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
MaverickTone wrote:
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.

-> 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?
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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Bunuel wrote:
Figure not drawn to scale.

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

Attachment:
main-qimg-5456f088c373424b9e0b766a7449d264.png

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
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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laxpro2001 wrote:
MaverickTone wrote:
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.

-> 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?

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.
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
As per the Menelaus Theorem,
(BC * ED * AF)/(BD*AE*FC)=1,

Assume BD=CD=x ; AE=ED=y ;
So, AF/FC=1/2.
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
VeritasPrepKarishma wrote:
Bunuel wrote:
Figure not drawn to scale.

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

Attachment:
main-qimg-5456f088c373424b9e0b766a7449d264.png

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.

Attachment:
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

Bunuel
I can't seem to find any good resources that help teach this concept - do you know of any?
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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Bunuel wrote:
Figure not drawn to scale.

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

Attachment:
The attachment main-qimg-5456f088c373424b9e0b766a7449d264.png is no longer available

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

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main-qimg-5456f088c373424b9e0b766a7449d264.png [ 15.84 KiB | Viewed 124948 times ]

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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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laxpro2001 wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:
Figure not drawn to scale.

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

Attachment:
main-qimg-5456f088c373424b9e0b766a7449d264.png

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.

Attachment:
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

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

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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
[quote="Bunuel"]Figure not drawn to scale.

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
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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Re: In triangle ABC, D is the mid-point of BC and E is the mid-point of AD [#permalink]
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