If median of the `Delta`ABC through A is perpendicular to BC, then which one of the following is correct ?

Options

A. tanA+tanB=0
B. 2tanA+tanB=0
C. tanA+2tanB=0
D. none of these

Answer

To solve the problem, we need to analyze the given condition that the median from vertex A to side BC of triangle ABC is perpendicular to BC. Let's go through the solution step by step:

### Step 1: Understand the Triangle and the Median
We are given triangle ABC, with A as one vertex and BC as the opposite side. The median from A to BC, which we will denote as AD, is perpendicular to BC. This means that AD bisects BC at point D and forms right angles with BC.

### Step 2: Set Up the Right Triangles
Since AD is a median, we have:
- BD = DC = 1/2 * BC (because D is the midpoint of BC).
- AD is perpendicular to BC, forming two right triangles: ABD and ACD.

### Step 3: Use Trigonometric Ratios
In triangle ABD:
- We can express tan(B) as:
\[ \tan(B) = \frac{AD}{BD} = \frac{AD}{\frac{1}{2}BC} = \frac{2AD}{BC} \]

In triangle ACD:
- Similarly, we can express tan(C) as:
\[ \tan(C) = \frac{AD}{DC} = \frac{AD}{\frac{1}{2}BC} = \frac{2AD}{BC} \]

### Step 4: Equate the Tangents
Since both expressions for tan(B) and tan(C) are equal, we have:
\[ \tan(B) = \tan(C) \]
This implies that angles B and C are equal:
\[ B = C \]

### Step 5: Analyze the Angles in Triangle ABC
From the triangle angle sum property, we know:
\[ A + B + C = 180^\circ \]
Since B = C, we can substitute:
\[ A + 2B = 180^\circ \]
Rearranging gives:
\[ A = 180^\circ - 2B \]

### Step 6: Use the Tangent Function
Taking the tangent of both sides:
\[ \tan(A) = \tan(180^\circ - 2B) \]
Using the property of tangent:
\[ \tan(180^\circ - \theta) = -\tan(\theta) \]
Thus:
\[ \tan(A) = -\tan(2B) \]

### Step 7: Final Expression
This leads us to the final equation:
\[ \tan(A) + \tan(2B) = 0 \]

### Conclusion
The correct answer is that if the median from A is perpendicular to BC, then:
\[ \tan(A) + 2\tan(B) = 0 \]