In a `DeltaABC`, AD is a median . The bisectors of `angleADBandangleADC` meet AB and AC at E and F respectively . If the ratio of AE : BE `= 2 : 3` , then find the ratio of EF : BC.

Options

A. `2 : 5`
B. `4 : 3`
C. `7 : 3`
D. `3 : 7`

Answer

To solve the problem step-by-step, we can follow these instructions:

### Step 1: Understand the Triangle and Given Information
We have triangle ABC with median AD. D is the midpoint of BC, meaning BD = DC. The angle bisectors of angles ADB and ADC intersect sides AB and AC at points E and F, respectively. We are given that the ratio AE:BE = 2:3.

**Hint:** Remember that a median divides the opposite side into two equal segments.

### Step 2: Assign Lengths Based on the Ratio
Since AE:BE = 2:3, we can assign lengths to these segments. Let AE = 2x and BE = 3x. Therefore, the total length of AB is:
\[ AB = AE + BE = 2x + 3x = 5x \]

**Hint:** Use variables to represent the lengths based on the given ratio for easier calculations.

### Step 3: Find Lengths of AD and BD
Since D is the midpoint of BC, we can denote BD = DC. Let BD = DC = y. Therefore, the total length of BC is:
\[ BC = BD + DC = y + y = 2y \]

**Hint:** Remember that the median divides the opposite side into two equal parts.

### Step 4: Use the Angle Bisector Theorem
According to the Angle Bisector Theorem, the angle bisector divides the opposite side in the ratio of the adjacent sides. For angle ADB, we have:
\[ \frac{AE}{BE} = \frac{AD}{BD} \]
Thus:
\[ \frac{2}{3} = \frac{AD}{3y} \]

From this, we can express AD in terms of y:
\[ AD = \frac{2}{3} \times 3y = 2y \]

**Hint:** The Angle Bisector Theorem is crucial for relating the segments created by angle bisectors.

### Step 5: Find Lengths of AF and CF
Similarly, for angle ADC, we have:
\[ \frac{AF}{CF} = \frac{AD}{DC} \]
Since DC = y:
\[ \frac{AF}{CF} = \frac{2y}{y} = 2 \]
This means:
\[ AF:CF = 2:1 \]

Let AF = 2k and CF = k. Therefore, the total length of AC is:
\[ AC = AF + CF = 2k + k = 3k \]

**Hint:** Use the same method of ratios to find the lengths of segments created by the angle bisector.

### Step 6: Apply Thales' Theorem
By Thales' theorem, we have:
\[ \frac{AE}{AB} = \frac{AF}{AC} = \frac{EF}{BC} \]

Substituting the lengths we found:
\[ \frac{2x}{5x} = \frac{2k}{3k} = \frac{EF}{2y} \]

This simplifies to:
\[ \frac{2}{5} = \frac{2}{3} = \frac{EF}{2y} \]

**Hint:** Thales' theorem relates the ratios of segments in similar triangles.

### Step 7: Solve for EF and BC Ratio
From the equation:
\[ \frac{EF}{2y} = \frac{2}{5} \]
We can express EF as:
\[ EF = \frac{2}{5} \times 2y = \frac{4y}{5} \]

Now, we know that:
\[ BC = 2y \]

Thus, the ratio EF:BC is:
\[ \frac{EF}{BC} = \frac{\frac{4y}{5}}{2y} = \frac{4}{10} = \frac{2}{5} \]

**Hint:** Simplifying the ratio will give you the final answer.

### Final Answer
The ratio EF:BC is \( 2:5 \).