Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let P be any point on the circumference of the circle with diameter AB. If AP meets the other circle at Q, then

In $$\triangle$$ AQC,

$$\angle$$ AQC $$= 90^\circ$$ ($$\because$$ Angle in a semi circle is $$90^\circ$$)

and in $$\triangle$$ APB,

$$\angle$$ APB $$= 90^\circ$$ ($$\because$$ Angle in a semi circle is $$90^\circ$$)

Comparing two triangles $$\triangle$$ APB and $$\triangle$$ AQC,

$$\angle$$ QAC $$= \angle PAB$$

$$\angle$$ AQC $$= \angle APB$$

$$\therefore \triangle APB = \triangle AQC$$

$$\therefore$$ QC // PB

Since we cannot prove that C is exactly midpoint of AB, QC $$= \frac{1}{2}$$PB cannot be proved