ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at P If O is the centre of - Maths - Circles

Question

ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect
at P If O is the centre of the circle and AB=DC, prove that:

i) triangle PAB is congruent to triangle PDC

ii) PA=PD and PC=PB

iii) AD||BC

Ashish Shukla · Accepted Answer

Dear Student,
<img src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ana_qa_image_58c291d329f20%20png">

Considering that the point P and O are same
i)In△PAB and △PCD PA =PC (radius of same circle
)
 PB=PD (radius of same circle)∠APB=∠CPD (opposite angle)
SO the △PAB ≅△PCD (SAS Rule)
Since △PAB ≅△PCD, corrosponding sides are equal,i
e
ii)PA=PD and PC=PBiii) now we are supposed to prove AD∥BC,In△PAD and △PBC PA =PC (radius of same circle
)
 PB=PD (radius of same circle)∠APD=∠BPC (opposite angle)
SO the △PAD ≅△PBC (SAS Rule)Now as the triangle are congrcongruent corresponding angle are i
e
∠PBC =∠PDA Since this angle are alternate angel between two line and are equal so the currosponding line AD and B arede parallel

Regards

ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at P. If O is the centre of the circle and AB=DC, prove that: i) triangle PAB is congruent to triangle PDC ii) PA=PD and PC=PB iii) AD||BC

ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at P. If O is the centre of the circle and AB=DC, prove that:

i) triangle PAB is congruent to triangle PDC

ii) PA=PD and PC=PB

iii) AD||BC