In a $\triangle$ ABC , points $D$ and $E$ are on the sides $BC$ and $AC$ , respectively. $BE$ and $AD$ intersect at point $T$ such that $AD:AT = 4:3$ , and $BE:BT = 5:4$. Point $F$ lies on $AC$ such that $DF$ is parallel to $BE$. Then, $BD:CD$ is

CAT 2025 Slot 2 QA Question 22

Multiple choice (+3 / −1) · Geometry · Triangles · Try it, then check the answer and solution below.

CAT 2025 Slot 2QAGeometry • TrianglesHard

In a

\triangle

ABC , points

D

and

E

are on the sides

BC

and

AC , respectively.

and

intersect at point

such that

AD:AT = 4:3 , and

BE:BT = 5:4

. Point

F

lies on

AC

such that

DF

is parallel to

BE

. Then,

BD:CD

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Official Convenor Answer Key: C

Official Correct Answer: C. Alright, so I've got this geometry problem from the CAT exam. But I'll take it slow and see if I can make sense of it. First, let me parse the problem again to make sure I understand all the details. We have triangle ABC. Points D and E are on sides BC and AC, respectively. BE and AD intersect at point T. The ratios given are AD:AT = 4:3 and BE:BT = 5:4. Then, there's a point F on AC such that DF is parallel to BE. We need to find the ratio BD:DC. Hmm, okay. So, we have triangle ABC with points D on BC and E on AC. Lines BE and AD intersect at T. The ratios of the segments are given, and then we have DF parallel to BE. I think I need to use some properties of triangles, maybe similar triangles or mass point geometry? I'm not entirely sure yet, but let's explore. Let me try to visualize the triangle. Let me sketch it mentally: triangle ABC, with D somewhere on BC, E on AC. BE is a line from B to E, and AD is a line from A to D. They intersect at T. The ratios are AD:AT = 4:3, which means that AT is 3 parts and TD is 1 part since AD is 4 parts. Similarly, BE:BT = 5:4, so BT is 4 parts and TE is 1 part. So, from AD:AT = 4:3, we can say that AT = (3/4)AD. Similarly, BT = (4/5)BE. Maybe I can use mass point geometry here because it deals with ratios along lines and points of intersection. Mass point assigns weights to the vertices so that the ratios of the sides can be determined. Let me recall how mass point works. If two cevians intersect, the masses can be assigned based on the ratios given. Let me assign masses to points A, B, and C such that the masses balance at point T. Since AD:AT = 4:3, the mass at A should be proportional to TD, which is 1 part, and the mass at D should be proportional to AT, which is 3 parts. Wait, no, mass points are a bit different. The masses are inversely proportional to the lengths. So, if AD:AT = 4:3, then the mass at A is proportional to TD, which is 1, and the mass at D is proportional to AT, which is 3. So, mass at A is 1, mass at D is 3. Similarly, for BE:BT = 5:4, the mass at B is proportional to TE, which is 1, and mass at E is proportional to BT, which is 4. So, mass at B is 1, mass at E is 4. Wait, but mass points require the masses to balance at the intersection point T. So, masses at A and D should balance at T, and masses at B and E should also balance at T. Therefore, the mass at T from AD's perspective is mass A + mass D = 1 + 3 = 4. Similarly, from BE's perspective, mass B + mass E = 1 + 4 = 5. But T should have the same mass from both perspectives, so I need to adjust the masses so that they are consistent. To make the masses at T equal, I can scale them. The mass from AD is 4, and from BE is 5. The least common multiple of 4 and 5 is 20. So, I can scale the masses on AD by 5 and the masses on BE by 4. That way, the mass at T becomes 20 from both sides. So, scaling AD: mass A becomes 1*5 = 5, mass D becomes 3*5 = 15. Scaling BE: mass B becomes 1*4 = 4, mass E becomes 4*4 = 16. Now, mass at A is 5, mass at B is 4, mass at D is 15, mass at E is 16. But mass at D is on BC, so masses at B and C should add up to mass at D. Similarly, mass at E is on AC, so masses at A and C should add up to mass at E. So, for point D on BC: mass B + mass C = mass D = 15. We know mass B is 4, so mass C = 15 - 4 = 11. For point E on AC: mass A + mass C = mass E = 16. We know mass A is 5, so mass C = 16 - 5 = 11. Perfect, that's consistent. So, mass at C is 11. Now, the masses at B and C are 4 and 11, respectively. Therefore, the ratio BD:DC is mass C : mass B, which is 11:4. Wait, is that right? Because in mass point, the ratio of the segments is inversely proportional to the masses. So, BD:DC = mass C : mass B = 11:4. Yes, So, BD:DC is 11:4, which is option C. But let me double-check to make sure I didn't make any mistakes. Let me go through the steps again. 1. Assign masses based on the given ratios. 2. AD:AT = 4:3, so masses at A and D are 1 and 3 respectively. 3. BE:BT = 5:4, so masses at B and E are 1 and 4 respectively. 4. To make the mass at T consistent, scale the masses so that mass T is 20 from both sides. 5. After scaling, masses at A=5, D=15, B=4, E=16. 6. Mass at D is 15, which is the sum of masses at B and C. So, mass C = 15 - 4 = 11. 7. Similarly, mass at E is 16, which is the sum of masses at A and C. So, mass C = 16 - 5 = 11. 8. Therefore, BD:DC = mass C : mass B = 11:4. Yes, that seems solid. I think I'm confident with this answer. The key was using mass point geometry to assign the masses correctly and then ensuring consistency by scaling. Once the masses are consistent, the ratio BD:DC is just the inverse of the masses at B and C. I also remember that when two cevians intersect, mass point can be a quick way to find the ratios without getting into coordinate geometry or similar triangles. It's a powerful tool for such problems. So, I think **Final Answer:** 11

Answer and solution

Answer: C) 11 : 4

Alright, so I've got this geometry problem from the CAT exam. But I'll take it slow and see if I can make sense of it.

First, let me parse the problem again to make sure I understand all the details. We have triangle ABC. Points D and E are on sides BC and AC, respectively. BE and AD intersect at point T. The ratios given are AD:AT = 4:3 and BE:BT = 5:4. Then, there's a point F on AC such that DF is parallel to BE. We need to find the ratio BD:DC.

Hmm, okay. So, we have triangle ABC with points D on BC and E on AC. Lines BE and AD intersect at T. The ratios of the segments are given, and then we have DF parallel to BE. I think I need to use some properties of triangles, maybe similar triangles or mass point geometry? I'm not entirely sure yet, but let's explore.

Let me try to visualize the triangle. Let me sketch it mentally: triangle ABC, with D somewhere on BC, E on AC. BE is a line from B to E, and AD is a line from A to D. They intersect at T. The ratios are AD:AT = 4:3, which means that AT is 3 parts and TD is 1 part since AD is 4 parts. Similarly, BE:BT = 5:4, so BT is 4 parts and TE is 1 part.

So, from AD:AT = 4:3, we can say that AT = (3/4)AD. Similarly, BT = (4/5)BE. Maybe I can use mass point geometry here because it deals with ratios along lines and points of intersection.

Mass point assigns weights to the vertices so that the ratios of the sides can be determined. Let me recall how mass point works. If two cevians intersect, the masses can be assigned based on the ratios given.

Let me assign masses to points A, B, and C such that the masses balance at point T. Since AD:AT = 4:3, the mass at A should be proportional to TD, which is 1 part, and the mass at D should be proportional to AT, which is 3 parts. Wait, no, mass points are a bit different. The masses are inversely proportional to the lengths.

So, if AD:AT = 4:3, then the mass at A is proportional to TD, which is 1, and the mass at D is proportional to AT, which is 3. So, mass at A is 1, mass at D is 3. Similarly, for BE:BT = 5:4, the mass at B is proportional to TE, which is 1, and mass at E is proportional to BT, which is 4. So, mass at B is 1, mass at E is 4.

Wait, but mass points require the masses to balance at the intersection point T. So, masses at A and D should balance at T, and masses at B and E should also balance at T. Therefore, the mass at T from AD's perspective is mass A + mass D = 1 + 3 = 4. Similarly, from BE's perspective, mass B + mass E = 1 + 4 = 5. But T should have the same mass from both perspectives, so I need to adjust the masses so that they are consistent.

To make the masses at T equal, I can scale them. The mass from AD is 4, and from BE is 5. The least common multiple of 4 and 5 is 20. So, I can scale the masses on AD by 5 and the masses on BE by 4. That way, the mass at T becomes 20 from both sides.

So, scaling AD: mass A becomes 15 = 5, mass D becomes 35 = 15. Scaling BE: mass B becomes 14 = 4, mass E becomes 44 = 16.

Now, mass at A is 5, mass at B is 4, mass at D is 15, mass at E is 16.

But mass at D is on BC, so masses at B and C should add up to mass at D. Similarly, mass at E is on AC, so masses at A and C should add up to mass at E.

So, for point D on BC: mass B + mass C = mass D = 15. We know mass B is 4, so mass C = 15 - 4 = 11.

For point E on AC: mass A + mass C = mass E = 16. We know mass A is 5, so mass C = 16 - 5 = 11. Perfect, that's consistent.

So, mass at C is 11. Now, the masses at B and C are 4 and 11, respectively. Therefore, the ratio BD:DC is mass C : mass B, which is 11:4.

Wait, is that right? Because in mass point, the ratio of the segments is inversely proportional to the masses. So, BD:DC = mass C : mass B = 11:4.

Yes, So, BD:DC is 11:4, which is option C.

But let me double-check to make sure I didn't make any mistakes. Let me go through the steps again.

Assign masses based on the given ratios.

AD:AT = 4:3, so masses at A and D are 1 and 3 respectively.

BE:BT = 5:4, so masses at B and E are 1 and 4 respectively.

To make the mass at T consistent, scale the masses so that mass T is 20 from both sides.

After scaling, masses at A=5, D=15, B=4, E=16.

Mass at D is 15, which is the sum of masses at B and C. So, mass C = 15 - 4 = 11.

Similarly, mass at E is 16, which is the sum of masses at A and C. So, mass C = 16 - 5 = 11.

Therefore, BD:DC = mass C : mass B = 11:4.

Yes, that seems solid. I think I'm confident with this answer. The key was using mass point geometry to assign the masses correctly and then ensuring consistency by scaling. Once the masses are consistent, the ratio BD:DC is just the inverse of the masses at B and C.

I also remember that when two cevians intersect, mass point can be a quick way to find the ratios without getting into coordinate geometry or similar triangles. It's a powerful tool for such problems.

So, I think

Final Answer: 11

CAT 2025 Slot 2 QA Question 22

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