`A B C D` is a cyclic quadrilateral whose diagonals `A Ca n dB D` intersect at `P` . If `A B=D C ,` Prove that : ` P A BcongP D C` `P A=P Da n dP C=P B` `A D B C`

Explanation

Let` P` and `Q` (center ) are same
In `trianglePAB` and `trianglePCD`
`PA=PC`......[radius]
`PB=PD`........[vertically opposite]
Therefore by `SAS`congravency
`angleAPB~=angleCPD`
By CPCT
`PA=PD`
`PC=PB`
In `triangle PAD` and `trianglePBC`
`PA=PC`.........[radius]
`PB=PD`.......[radius]
`angle APD=angleBPC`.........[vertical opposite]
By `SAS` congranency
`triangle PADcongtrianglePBC`
By CPCT
`angle PBC=anglePDC`....[alternates angles between two lines and are equal ]
lines are equal `AD=BC`