The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at the point P. Then, it is always true that
As we know that a cyclic quadrilateral can be inscribed into a circle, Hence in triangle APB and in triangle CPD.
$$\angle PAB = \angle PDC$$ (same sector angles)
$$\angle PCD = \angle PBA$$ (same sector angles)
Hence third angle will also be equal and they will be similar triangles.
So $$\frac{AP}{PD} = \frac{BP}{PC}$$
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