If A(1,5), B (-2,1) and C(4,1) be the vertices of ΔABC and the internal bisector of ∠A meets BC and D, find AD.

Given: A(1, 5), B(-2, 1) and C(4,1) are the vertices of ΔABC

Using angle bisector theorem, which states that:

The ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:{displaystyle {frac {|BD|}{|DC|}}={frac {|AB|}{|AC|}},}

⇒ BD = DC

⇒ D is the midpoint of BC

So, the coordinates of D are:

D = (1, 1)

Now, AD = √(x₂ – x₁)² + (y₂ – y₁)²

= √(1 – 1)² + {5 – 1}²

= √(0)² + (4)²

= √16

= 4 units

Hence, AD = 4 units