Question

$\triangle$ABC is obtuse angled at B and points D and E are taken on sides AC and BC respectively such that

AD: DC = 2 : 3 and BE : EC = 1 : 2. If DE produced meet AB produced at F, then -

• A. AF : BF = 4 : 3
• B. AF : BF = 3 : 1
• C. DE : EF = 1 : 5
• D. DE : EF = 3 : 4

Answer 1

Answer: (A), (C)

(A), (C)

Answer 2

Let the vertices of the triangle be A, B, and C. Let B be the origin, so $\vec{B} = \mathbf{0}$. Let $\vec{A} = \mathbf{a}$ and $\vec{C} = \mathbf{c}$.

Point D is on AC such that AD:DC = 2:3. Using the section formula, the position vector of D is $\vec{D} = \frac{3\vec{A} + 2\vec{C}}{3+2} = \frac{3\mathbf{a} + 2\mathbf{c}}{5}$.

Point E is on BC such that BE:EC = 1:2. Since B is the origin, $\vec{BE} = \frac{1}{3}\vec{BC}$. So, the position vector of E is $\vec{E} = \frac{1}{3}\vec{C} = \frac{1}{3}\mathbf{c}$.

The line segment DE is produced to meet AB produced at F. This means F lies on the line DE and also on the line AB. Since F lies on the line DE, its position vector $\vec{F}$ can be expressed as a linear combination of $\vec{D}$ and $\vec{E}$:

$\vec{F} = (1-s)\vec{D} + s\vec{E}$ for some scalar s.

Substitute the expressions for $\vec{D}$ and $\vec{E}$:

$\vec{F} = (1-s)\left(\frac{3\mathbf{a} + 2\mathbf{c}}{5}\right) + s\left(\frac{1}{3}\mathbf{c}\right)$

$\vec{F} = \frac{3(1-s)}{5}\mathbf{a} + \frac{2(1-s)}{5}\mathbf{c} + \frac{s}{3}\mathbf{c}$

$\vec{F} = \frac{3(1-s)}{5}\mathbf{a} + \left(\frac{2(1-s)}{5} + \frac{s}{3}\right)\mathbf{c}$

$\vec{F} = \frac{3(1-s)}{5}\mathbf{a} + \frac{6-s}{15}\mathbf{c}$

Since F lies on the line AB produced, F lies on the line passing through B (origin) and A (vector $\mathbf{a}$). Thus, $\vec{F}$ must be a multiple of $\vec{A}$ (vector $\mathbf{a}$).

$\vec{F} = k\mathbf{a}$ for some scalar k.

Comparing the two expressions for $\vec{F}$:

$k\mathbf{a} = \frac{3(1-s)}{5}\mathbf{a} + \frac{6-s}{15}\mathbf{c}$

Since $\mathbf{a}$ and $\mathbf{c}$ are not parallel (as they form a triangle), the coefficients of $\mathbf{a}$ and $\mathbf{c}$ on both sides must be equal.

Coefficient of $\mathbf{c}$: $0 = \frac{6-s}{15} \implies s = 6$.

Coefficient of $\mathbf{a}$: $k = \frac{3(1-s)}{5} = \frac{3(1-6)}{5} = \frac{3(-5)}{5} = -3$.

So, $\vec{F} = -3\mathbf{a}$.

Since $\vec{A} = \mathbf{a}$ and $\vec{B} = \mathbf{0}$, $\vec{BA} = \vec{A} - \vec{B} = \mathbf{a}$.

$\vec{BF} = \vec{F} - \vec{B} = -3\mathbf{a} - \mathbf{0} = -3\mathbf{a}$.

So, $\vec{BF} = -3\vec{BA}$.

Since $\vec{BA} = -\vec{AB}$, we have $\vec{BF} = -3(-\vec{AB}) = 3\vec{AB}$.

This implies that F is on the line AB and $\vec{BF}$ is in the same direction as $\vec{AB}$. The order of points must be B, A, F.

The magnitudes are related by $BF = 3 AB$.

Since B, A, F are collinear in that order, $BF = BA + AF$.

$3 AB = AB + AF$.

$AF = 3 AB - AB = 2 AB$.

The ratio AF : BF = $2 AB : 3 AB = 2 : 3$.

Let's recheck the Menelaus' theorem application.

Consider $\triangle$ABC and the transversal line FDE. The line intersects AB produced at F, BC at E, and AC at D.

According to Menelaus' Theorem:

$(\frac{AF}{FB}) \cdot (\frac{BE}{EC}) \cdot (\frac{CD}{DA}) = 1$.

We are given BE:EC = 1:2, so $\frac{BE}{EC} = \frac{1}{2}$.

We are given AD:DC = 2:3, so $\frac{AD}{DC} = \frac{2}{3}$. This means $\frac{CD}{DA} = \frac{3}{2}$.

Substitute these values into the Menelaus' equation:

$(\frac{AF}{FB}) \cdot (\frac{1}{2}) \cdot (\frac{3}{2}) = 1$

$(\frac{AF}{FB}) \cdot \frac{3}{4} = 1$

$\frac{AF}{FB} = \frac{4}{3}$.

So, AF : BF = 4 : 3. This matches option (A).

Now let's find the ratio DE : EF.

We used the scalar $s=6$ in the equation $\vec{F} = (1-s)\vec{D} + s\vec{E}$.

$\vec{F} = (1-6)\vec{D} + 6\vec{E}$

$\vec{F} = -5\vec{D} + 6\vec{E}$.

Rearranging this equation to relate vectors from E:

$\vec{F} - \vec{E} = -5\vec{D} + 6\vec{E} - \vec{E}$

$\vec{EF} = -5\vec{D} + 5\vec{E}$

$\vec{EF} = -5(\vec{D} - \vec{E})$

$\vec{EF} = -5\vec{ED}$.

This equation shows that the vector $\vec{EF}$ is 5 times the magnitude of the vector $\vec{ED}$ and points in the opposite direction.

Since $\vec{EF}$ and $\vec{ED}$ are opposite, E must lie between D and F.

The order of points is D, E, F.

The magnitude relation is $EF = 5 ED$.

The ratio DE : EF = ED : EF = $ED : 5 ED = 1 : 5$.

This matches option (C).

Let's verify the order of points D, E, F using the scalar s.

$\vec{F} = (1-s)\vec{D} + s\vec{E}$. If $0 < s < 1$, E is between D and F. If $s < 0$, D is between E and F. If $s > 1$, E is between D and F.

The line segment DE is produced to F. This means E is between D and F or D is between E and F.

The equation $\vec{F} = (1-s)\vec{D} + s\vec{E}$ implies that F lies on the line passing through D and E.

We found $s=6$.

$\vec{F} = (1-6)\vec{D} + 6\vec{E} = -5\vec{D} + 6\vec{E}$.

Consider the vector $\vec{DF}$. $\vec{DF} = \vec{F} - \vec{D} = (-5\vec{D} + 6\vec{E}) - \vec{D} = -6\vec{D} + 6\vec{E} = 6(\vec{E} - \vec{D}) = 6\vec{DE}$.

This means the vector $\vec{DF}$ is 6 times the vector $\vec{DE}$ and in the same direction.

The points D, E, F are collinear and ordered as D, E, F.

$DF = 6 DE$.

Since D, E, F are in order, $DF = DE + EF$.

$6 DE = DE + EF$.

$EF = 6 DE - DE = 5 DE$.

The ratio DE : EF = DE : 5 DE = 1 : 5.

This confirms option (C).