A triangle ABC with angle B =90 degrees. Taking AB as diameter a circle has been drawn intersecting AC at a point P. Prove that tangent drawn at a point P bisect BC

To prove that the tangent drawn at point P bisects BC, we can use the properties of a circle and a right triangle. Here are the steps:

1. Since AB is the diameter of the circle and angle B is 90 degrees, we can say that angle APB is 90 degrees because the angle in a semi-circle is a right angle.

2. Now, let's draw a line from P to B and call it PB. Since angle APB is 90 degrees, triangle APB is a right triangle.

3. Let's draw a tangent at point P and call it PT. Since PT is a tangent to the circle at P, it is perpendicular to the radius at the point of contact. So, angle APB = angle PTB = 90 degrees.

4. Now, let's draw a line from P to C and call it PC. Since angle B is 90 degrees and angle APB = angle PTB = 90 degrees, we can say that triangle PBC is similar to triangle PTB by AA similarity criterion (two triangles are similar if they have two corresponding angles equal).

5. In similar triangles, corresponding sides are in the same ratio. So, PB/PT = BC/PC.

6. But PB = PT (tangent from an external point to a circle are equal), so BC = PC. This means that point P bisects BC.

Therefore, the tangent drawn at point P bisects BC.