Q: 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
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Aman verma
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Q: 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
Last edited by Aman verma on Thu Nov 14, 2013 7:00 am, edited 1 time in total.
800. Arjun's-Bird-Eye
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Aman verma
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theCodeToGMAT
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I believe there is some missing information in this question because diagram says that it is not drawn to scale and the condition "D is the mid-point of BC and E is the mid-point of AD. BF passes through E" can also be possible for an equilateral triangle and in that case AF:FC = 1:1..
Are you sure of the question? What is the source?
Are you sure of the question? What is the source?
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Mathsbuddy
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The correct answer is 1:2.
Firstly, it doesn't matter what kind of triangle it is because it is all about relative proportion.
Therefore, we can redraw it as a right angle triangle if we want. It won't change the answer.
If this new triangle sits on x-y axes, imagine vertex B at the origin, vertex A somewhere up the y-axis at (0,a) and vertex C somewhere along the x-axis at (c,0). A sketch will help visualise this.
By looking at similar triangles, we see that point E is at (c/4,a/2), which means it has a gradient (slope) of 2a/c
If an extra line is drawn passing through point A, parallel to BE, it has the same gradient. From which we can calculate that it intercepts the x-axis at new point G(-c/2,0).
Triangle GAC is similar to BFC, therefore the ratio AF:FC = GB:BC = 1:2
Firstly, it doesn't matter what kind of triangle it is because it is all about relative proportion.
Therefore, we can redraw it as a right angle triangle if we want. It won't change the answer.
If this new triangle sits on x-y axes, imagine vertex B at the origin, vertex A somewhere up the y-axis at (0,a) and vertex C somewhere along the x-axis at (c,0). A sketch will help visualise this.
By looking at similar triangles, we see that point E is at (c/4,a/2), which means it has a gradient (slope) of 2a/c
If an extra line is drawn passing through point A, parallel to BE, it has the same gradient. From which we can calculate that it intercepts the x-axis at new point G(-c/2,0).
Triangle GAC is similar to BFC, therefore the ratio AF:FC = GB:BC = 1:2
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Brent@GMATPrepNow
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VERY tricky question. I'd say it's skirting the 800 level of difficulty!Aman verma wrote:
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
Here's a solution with some diagrams.
First, we'll add the given information.
Next, from point D let's add a line that's PARALLEL to BF
This addition of a parallel line allows us to see a variety of equivalent angles (and similar triangles!)
First, let's examine these two similar triangles.
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.
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
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Sun Apr 29, 2018 5:39 am, edited 1 time in total.
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The triangle can be of ANY FORM that satisfies the given constraints.
Redraw ABC as a 6-8-10 triangle such that AC=6, BC=8 and AB=10:
∆ABC:
Since D is the midpoint of BC, BD=DC=4.
∆ACD:
When a line is drawn from the midpoint of the hypotenuse to a leg, the leg is BISECTED.
Thus, EH bisects CD, and EG bisects AC.
Result:
DH=2, implying that BH=6.
CG=3, implying that EH=3.
∆BEH and ∆BFC:
Since EH is parallel to CF, ∆BEH and ∆BFC have the same combination of angles and thus are SIMILAR.
Corresponding sides of similar triangles are in the SAME RATIO.
Side BH in ∆BEH corresponds to side BC in ∆BFC.
Side EH in ∆BEH corresponds to side FC in ∆BFC.
Thus:
BH/BC = EH/FC
6/8 = 3/FC
6(FC) = 24
FC= 4.
Side AC:
Since AC=6 and FC=4, AF=2.
Thus:
AF:FC = 2:4 = 1:2.
The correct answer is B.
Redraw ABC as a 6-8-10 triangle such that AC=6, BC=8 and AB=10:
∆ABC:
Since D is the midpoint of BC, BD=DC=4.
∆ACD:
When a line is drawn from the midpoint of the hypotenuse to a leg, the leg is BISECTED.
Thus, EH bisects CD, and EG bisects AC.
Result:
DH=2, implying that BH=6.
CG=3, implying that EH=3.
∆BEH and ∆BFC:
Since EH is parallel to CF, ∆BEH and ∆BFC have the same combination of angles and thus are SIMILAR.
Corresponding sides of similar triangles are in the SAME RATIO.
Side BH in ∆BEH corresponds to side BC in ∆BFC.
Side EH in ∆BEH corresponds to side FC in ∆BFC.
Thus:
BH/BC = EH/FC
6/8 = 3/FC
6(FC) = 24
FC= 4.
Side AC:
Since AC=6 and FC=4, AF=2.
Thus:
AF:FC = 2:4 = 1:2.
The correct answer is B.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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Aman verma
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Hi all !
Many thanks to the GMAT instructors and mathbuddy for making the concept clear. I had the net answer (B) but I didn't had the detailed solution to this problem. Thank you for explaining and clarifying the underlying concepts.
Now the crux of the problem depends on realizing the fact that the solution relies on the ratio of magnitudes than the shape of the figure. I had devised one solution using similarity of triangles concept but I don't know how much that was correct. My solution:- Triangles AFE~ACD~BFC then:
AE/ED=BD/BC=AF/FC=1/2. If we look at Triangle ADC in reverse it appears similar to Triangle AEF,
but now I realize that reasoning might be wrong. So I would like to know how much my solution seems plausible.
Many thanks to the GMAT instructors and mathbuddy for making the concept clear. I had the net answer (B) but I didn't had the detailed solution to this problem. Thank you for explaining and clarifying the underlying concepts.
Now the crux of the problem depends on realizing the fact that the solution relies on the ratio of magnitudes than the shape of the figure. I had devised one solution using similarity of triangles concept but I don't know how much that was correct. My solution:- Triangles AFE~ACD~BFC then:
AE/ED=BD/BC=AF/FC=1/2. If we look at Triangle ADC in reverse it appears similar to Triangle AEF,
but now I realize that reasoning might be wrong. So I would like to know how much my solution seems plausible.
800. Arjun's-Bird-Eye
