CAT 2025 Slot 2 · Quant · Question 21

AD:AT=4:3 means T divides AD with AT:TD=3:1, and BE:BT=5:4 means BT:TE=4:1.

Place masses at A, B and C so that T balances both cevians, a mass being inversely proportional to its distance from the balance point.

Along AD, AT:TD=3:1 needs {m}_{A}\times 3={m}_{D}\times 1, so {m}_{D}=3{m}_{A}.

Along BE, BT:TE=4:1 needs {m}_{B}\times 4={m}_{E}\times 1, so {m}_{E}=4{m}_{B}.

Since D lies on BC, {m}_{D}={m}_{B}+{m}_{C}; since E lies on AC, {m}_{E}={m}_{A}+{m}_{C}.

Writing {m}_{A}=a, {m}_{B}=b and {m}_{C}=c, these become 3a=b+c and 4b=a+c.

From the first, c=3a-b; substituting into the second, 4b=4a-b, so 5b=4a and b=\frac{4a}{5}.

Then c=3a-\frac{4a}{5}=\frac{11a}{5}.

D divides BC inversely to the masses at B and C, so BD:DC={m}_{C}:{m}_{B}=\frac{11a}{5}:\frac{4a}{5}=11:4.

The point F with DF\parallel BE is the textbook route — in triangle CBE it gives CD:DB=CF:FE — but the answer does not depend on it.

Shortcut. Mass points turn this into two linear equations and skip similar triangles altogether. The pattern is worth memorising: a cevian split of m:n at the intersection puts masses in the ratio n:m at its two ends, and a point on a side carries the sum of the masses at that side's endpoints.