Question

In the figure, chords $BA$ and $DC$ of a circle meet at $P$. Prove that
$PA\times PB=PC\times PD$

Answer 1

To prove ,$PA\times PB=PC\times PD$

Consider $△PCA$ and $PBD$

$\angle PCA=\angle PBD$ (Angle in the same segment)

$\angle APC=\angle BPD$ (Vertically opposite angles)

$△PCA\sim △PBD$ (AA Similarity\right)

P lies outside the circle,

$\angle PCA+\angle CAB={180}^0$ (Linear pair)

$\angle CAB+\angle PDB={180}^0$ (Opposite angles of a cyclic)

$\angle PAC=\angle PDB$

In $△PCA$ and $△PBD$

$\angle PAC=\angle PDB$ (proved above)

$\angle APC=\angle DPB$ (common)

$\angle PCA\sim \angle PBD$ (AA similarity)

Hence

$\frac{PA}{PD}=\frac{PC}{PB}$

$PA\times PB=PC\times PD$

Hence proved