ABC is an equilateral triangle .If AD bisects angle A, Prove that AD is perpendicular bisector of BC .

Answer

To prove that AD is the perpendicular bisector of BC in an equilateral triangle ABC where AD bisects angle A, we will follow these steps:

### Step 1: Understand the properties of an equilateral triangle
In an equilateral triangle, all sides are equal, and all angles are equal to 60 degrees.

### Step 2: Identify the angles
Since AD bisects angle A, we have:
\[ \angle BAD = \angle DAC \]

### Step 3: Establish congruence of triangles
Consider triangles ABD and ACD. We know:
- \( AB = AC \) (sides of the equilateral triangle are equal)
- \( AD \) is common to both triangles
- \( \angle BAD = \angle DAC \) (since AD bisects angle A)

By the Side-Angle-Side (SAS) congruence criterion, we can conclude:
\[ \triangle ABD \cong \triangle ACD \]

### Step 4: Conclude properties from congruence
Since the triangles are congruent, we have:
- \( BD = DC \) (corresponding parts of congruent triangles are equal)

This shows that D is the midpoint of BC, hence AD bisects BC.

### Step 5: Show that AD is perpendicular to BC
From the congruence of triangles ABD and ACD, we also have:
\[ \angle ADB = \angle ADC \]

Since these two angles form a linear pair (they are adjacent angles on a straight line), we can write:
\[ \angle ADB + \angle ADC = 180^\circ \]

Let \( \angle ADB = \angle ADC = x \). Then:
\[ x + x = 180^\circ \]
\[ 2x = 180^\circ \]
\[ x = 90^\circ \]

Thus, \( \angle ADB = 90^\circ \), which means AD is perpendicular to BC.

### Conclusion
We have shown that AD is both the bisector of BC and perpendicular to it, proving that AD is the perpendicular bisector of BC.

---